Entry-Level段子手的日常

Missing Number

2016-04-15T08:41:26.000Z

https://leetcode.com/problems/missing-number/

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity.

Solution 1: Binary Search

Time complexity: O(NlogN); Space Complexity O(1)

public class Solution {
    public int missingNumber(int[] nums) {
        Arrays.sort(nums);
        if (nums.length == 1) {
            return nums[0] == 0 ? 1 : 0;
        }
        
        int end = nums.length - 1;
        int start = 0;
        
        while (start + 1 < end) {
            int mid = start + (end -start) / 2;
            if (nums[mid] == mid) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if (nums[end] == end) {
            return nums[end] + 1;
        } else if (nums[end - 1] == end - 1){
            return end;
        } else {
            return end - 1;
        }
    }
}

Solution 2: Extra O(N) Space

Time complexity: O(N); Space Complexity O(N)

public class Solution {
    public int missingNumber(int[] nums) {
        int[] count = new int[nums.length + 1];
        for (int i = 0; i < nums.length; i++) {
            count[nums[i]]++;
        }
        for (int i = 0; i < count.length; i++) {
            if (count[i] == 0) {
                return i;
            }
        }
        return -1;
    }
}

Solution 3: Bit Manipulation

reference: http://believerw.github.io/2016/04/14/Bit-Manipulation/

Time complexity: O(N); Space Complexity O(1)

public class Solution {
    public int missingNumber(int[] nums) {
        int result = 0;
        for (int i = 0; i <= nums.length; i++) {
            result ^= i;
        }
        for (int i = 0; i < nums.length; i++) {
            result ^= nums[i];
        }
        return result;
    }
}

Solution 4: Method Without Bit Manipulation

reference: https://leetcode.com/discuss/96842/1ms-java-solution-without-bit-manipulation

Time complexity: O(N); Space Complexity O(1)

public class Solution {
    public int missingNumber(int[] nums) {
        int sum = 0, i = 0;

        for (; i < nums.length; i++)
            sum = sum + i - nums[i];

        return (sum + i);
    }
}

Bit Manipulation

2016-04-14T05:06:07.000Z

reference: http://www.cnblogs.com/maples7/p/4324844.html

Convert a decimal number into binary number with bit manipulation

for clause judges the each bit in a 16-bit binary number.
In the for loop, we use & and <<. The effect of << operation is like this: when i = 15, (1 << i) will convert 0000000000000001 into 1000000000000000; and when i = 10, the result will be 0000010000000000. So the result of num &（1 << i is comparing the ith digit (i is from 0 to 15) in num with (1 << i). Only when ith digit in num is 1, can num &（1 << i） be true.

public String bitManipulation(int num) {
    StringBuilder result = new StringBuilder();
    
    for (int i = 15; i >= 0; i--) {
        if ((num & (1 << i)) != 0) {
            result.append(1);
        } else {
            result.append(0);
        }
    }
    return result.toString();
}

LeetCode Example 1 - Number of 1 Bits

https://leetcode.com/problems/number-of-1-bits/

Write a function that takes an unsigned integer and returns the number of 1 bits it has (also known as the Hamming weight).

For example, the 32-bit integer 11 has binary representation 00000000000000000000000000001011, so the function should return 3.

Solution 1

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;
        for (int i = 0; i < 32; i++) {
            if ((n & (1 << i)) != 0) {
                count ++;
            }
        }
        
        return count;
    }
}

Solution 2 O(logN)

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;
        
        while(n > 0) {
            count += (n & 1);
            n >> 1
        }
        
        return count;
    }
}

The time complexity is O(logN), Since for each >> operation, N will be N / 2.

Solution 3 O(countOnes(N))

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;
        
        while(n > 0) {
            count ++;
            n &= n - 1
        }
        
        return count;
    }
}

For an arbitrary n, we can convert it to a binary number which is xxx...xxx1000....000. We do not take of how many x there are and how many 1s in the sequence of x, we just focus on 1000...000.

Since n is xxx...xxx1000....000, then n - 1 must be xxx...xxx0111....111 with the same numbers of digits after xxx...xxx. When we use (n & (n - 1)) operation, we will get xxx...xxx0000....000, which means we remove the last 1 in n.

LeetCode Example 2 - Reverse Bits

https://leetcode.com/problems/reverse-bits/

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

Follow up:
If this function is called many times, how would you optimize it?

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int result = 0;

        for (int i = 0; i < 32; i++) {
            result = (result << 1) | (n & 1);
            n = n >> 1;
        }
        return result;
    }
}

LeetCode Solution - Single Number

2016-04-14T02:50:59.000Z

https://leetcode.com/problems/single-number/

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Single Number I

Solution 1: With HashSet

Time Complexity: O(N^2), Space Complexity: O(N);

public class Solution {
    public int singleNumber(int[] nums) {
        Set<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (set.contains(nums[i])) {
                set.remove(nums[i]);
            } else {
                set.add(nums[i]);
            }
        }
        
        for (int i = 0; i < nums.length; i++) {
            if (set.contains(nums[i])) {
                return nums[i];
            }
        }
        return -1;
    }
}

Solution 2: With Sort Algorithm

Time Complexity: O(NlogN), Space Complexity: O(1);

public class Solution {
    public int singleNumber(int[] nums) {
    Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if (i == 0) {
                if (nums.length == 1 || nums[0] != nums[1]) {
                    return nums[0];
                }
            } else if(i == nums.length - 1) {
                if (nums[i] != nums[i - 1]) {
                    return nums[i];
                }
            } else {
                if (nums[i] != nums[i - 1] && nums[i] != nums[i + 1]) {
                    return nums[i];
                }
            }
        }
        return -1;
    }
}

Solution 3: With XOR Operation

reference: https://leetcode.com/discuss/94594/java-solution-time-complexity-and-space-complexity-using-xor
refernce: https://leetcode.com/discuss/85396/how-can-xor-keep-many-numbers

Time Complexity: O(N), Space Complexity: O(1);
The idea hinges on 3 properties of xor. (1) that its a commutative operation (i.e. a xor b = b xor a). (2) that something xor itself is 0. So a xor a = 0. And (3) 0 xor a = a. These three properties mean that

a xor b xor a = a xor a xor b = 0 xor b = b.

Thus it doesn’t matter the order of the numbers. If something only occurs once it won’t get negated.

public class Solution {
    public int singleNumber(int[] nums) {
        int res = 0;
        for(int num : nums) {
            res ^= num;
        }
        return res;
    }
}

Single Number II

https://leetcode.com/problems/single-number-ii/
Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity.

Solution 1: With HashMap

public class Solution {
    public int singleNumber(int[] nums) {
        Map<Integer, Integer> myMap = new HashMap<Integer, Integer>();

        //Filling up the map
        for(int i=0;i<nums.length;i++){
            if(!myMap.containsKey(nums[i])){
                myMap.put(nums[i], 1);
            }
            else{
                int temp = myMap.get(nums[i]);
                temp++;
                myMap.put(nums[i], temp);
            }
        }

        // Iterating over the map and returning the number with 
        // the value pair having a value less than 3

        Iterator it = myMap.entrySet().iterator();
        while(it.hasNext()){
            Map.Entry pair = (Map.Entry)it.next();
            int tmpValue = (int)pair.getValue();
            int tmpKey = (int)pair.getKey();
            if(tmpValue<3){
                return tmpKey;
            }
        }
        return -1;
    }
}

Solution 2: With Bit Manipulation

For a 32-bit integer, if this integer appears three times in the input array, each digit of this integer will apppears three times. While if it only appears once in the input array, each digit of this integer will appeaars only once.

public class Solution {
    public int singleNumber(int[] nums) {
        int[] count = new int[32];
        int result = 0;
        
        for (int j = 0; j < 32; j++) {
            for (int i = 0; i < nums.length; i++) {
                if ((nums[i] & (1 << j)) != 0) {
                    count[j] ++;
                }
            }
            if (count[j] % 3 != 0) {
                result += (1 << j);
            }
        }
        
        return result;
    }
}

Single Number III

https://leetcode.com/problems/single-number-iii/
Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

The order of the result is not important. So in the above example, [5, 3] is also correct.

Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

Solution 1: With HashMap

https://leetcode.com/discuss/95580/java-easy-to-understand-solution-with-map

public class Solution {
    public int[] singleNumber(int[] nums) {

        Map<Integer, Integer> myMap = new HashMap<Integer, Integer>();

        //Filling up the map
        for(int i=0;i<nums.length;i++){
            if(!myMap.containsKey(nums[i])){
                myMap.put(nums[i], 1);
            }
            else{
                int temp = myMap.get(nums[i]);
                temp++;
                myMap.put(nums[i], temp);
            }
        }

        //Iterating over the map and returning the number with the value pair having a value less than 3
        List<Integer> myList = new ArrayList<Integer>();
        Iterator it = myMap.entrySet().iterator();
        while(it.hasNext()){
            Map.Entry pair = (Map.Entry)it.next();
            int tmpValue = (int)pair.getValue();
            int tmpKey = (int)pair.getKey();
            if(tmpValue<2){
                myList.add(tmpKey);
            }
        }
        int[] result = new int[2];
        for(int i=0;i<myList.size();i++){
            result[i] = myList.get(i);
        }
        return result;
    }
}

Solution 2: With XOR Operation

Use XOR for all Integer in the input array, the result is the XOR result between two number like result = 3^5. We know that 3 is not equal to 5 so that 3^5 is not eqaul to 0 and there must be one digit in 3 and 5 are not equal, for example, this digit in 3 is 0 while is 1 in 5. (if not, 3^5 will be 0)

With this property, we can divide the input array inot two groups: one group contains all integer with 0 in that digit, and the another contain all integer with 1 in that digit.

Use XOR for these two groups separately, we will get 3 and 5 separately. And then with 3^result = 5 and 5^result = 3 we will find the answer of this question.

public class Solution {
    public int[] singleNumber(int[] nums) {
        int result = 0;
        for (int i = 0; i < nums.length; i++) {
            result ^= nums[i];
        }
        
        int pos = 0;
        for(int i = 0; i < 15; i++) {
            if ((result & (1 << i)) != 0) {
                pos = i;
            }
        }
        int groupA = 0, groupB = 0;
        for (int i = 0; i < nums.length; i++) {
            if ((nums[i] & (1 << pos)) != 0) {
                groupA ^= nums[i];
            } else {
                groupB ^= nums[i];
            }
        }
        
        int[] ans = new int[2];
        ans[0] = result ^ groupA;
        ans[1] = result ^ groupB;
        
        return ans;
    }
}

More Reference:

http://blog.sina.com.cn/s/blog_77dd9d490102vwew.html
http://blog.csdn.net/lcj_cjfykx/article/details/48531073
http://believerw.github.io/2016/04/14/Bit-Manipulation/#more

LintCode Solution - Binary Representation

2016-04-10T05:33:14.000Z

http://www.lintcode.com/en/problem/binary-representation/

Given a (decimal - e.g. 3.72) number that is passed in as a string, return the binary representation that is passed in as a string. If the fractional part of the number can not be represented accurately in binary with at most 32 characters, return ERROR.
Example
For n = "3.72", return "ERROR".
For n = "3.5", return "11.1".

Solution

public class Solution {
    /**
     *@param n: Given a decimal number that is passed in as a string
     *@return: A string
     */
    public String binaryRepresentation(String n) {
        // write your code here
        int integer = Integer.parseInt(n.substring(0,n.indexOf('.')));
        double fraction = Double.parseDouble(n.substring(n.indexOf('.')));

        StringBuilder result = new StringBuilder();
        boolean flag = false;
        
        for (int i = 0; i < 32; i++) {
            if (!(fraction > 0.0)) {
                flag = true;
                break;
            }
            int curry;
            fraction = fraction * 2;
            curry = (int) fraction / 1;
            fraction = fraction - curry;
            result.append(String.valueOf(curry));
        }
        
        if (flag) {
            StringBuilder temp = new StringBuilder();
            while (integer != 0) {
                temp.append(integer % 2);
                integer = integer / 2;
            }
            return binaryRepresentHelper(temp, result);
        } else {
            return "ERROR";
        }
    }
    private String binaryRepresentHelper(StringBuilder integer, StringBuilder fraction) {
        if (fraction.length() == 0 && integer.length() == 0) {
            return "0";
        }else if (fraction.length() == 0) {
            return integer.reverse().toString();
        } else if (integer.length() == 0) {
            return "0." + fraction.toString();
        } else {
            return integer.reverse().toString() + "." + fraction.toString();
        }
    }
}

ATTENTION

float is not available here since Java has taken the first 6 numbers after the point and then rounded up the rest. float is less precise than a double.
int integer = Integer.parseInt(n.substring(0,n.indexOf('.')));
double fraction = Double.parseDouble(n.substring(n.indexOf('.')));

LintCode Solution - Anagrams

2016-04-10T05:01:13.000Z

http://www.lintcode.com/en/problem/anagrams/

Given an array of strings, return all groups of strings that are anagrams.

Example
Given ["lint", "intl", "inlt", "code"], return ["lint", "inlt", "intl"].
Given ["ab", "ba", "cd", "dc", "e"], return ["ab", "ba", "cd", "dc"].

Solution 1

Thinking:

Using a HashTable to store anagrams (Using the smallest string as a Key)
- For example: string ba‘s key is ab, and its value is itself, ba.
For each string in the input array, calculate its key and store (key, value) into the hashtable with mpa.get(key).add(value). In this case, group anagrams will store in the fomation like key: [abc], value [abc, acb, bca, cba].

public class Solution {
    /**
     * @param strs: A list of strings
     * @return: A list of strings
     */
    public List<String> anagrams(String[] strs) {
        List<String> result = new ArrayList<String>();
        HashMap<String, ArrayList<String>> map = new HashMap<String, ArrayList<String>>();

        for (String str : strs) {
            char[] key = str.toCharArray();
            Arrays.sort(key);
            
            String Key = String.valueOf(key);
            if (!map.containsKey(Key)) {
                map.put(Key, new ArrayList<String>());
            }

            map.get(Key).add(str);
        }

        for (ArrayList<String> tmp : map.values()) {
            if (tmp.size() > 1) {
                result.addAll(tmp);
            }
        }

        return result;
    }
}

Solution 2

Thinking: The key in hashtable for this solution is calculated by getHash()

/**
 * 本代码由九章算法编辑提供。没有版权欢迎转发。
 * - 九章算法致力于帮助更多中国人找到好的工作，教师团队均来自硅谷和国内的一线大公司在职工程师。
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 * - 更多详情请见官方网站：http://www.jiuzhang.com/
 */

public class Solution {
    private int getHash(int[] count) {
        int hash = 0;
        int a = 378551;
        int b = 63689;
        for (int num : count) {
            hash = hash * a + num;
            a = a * b;
        }
        return hash;
    }
    
    public ArrayList<String> anagrams(String[] strs) {
        ArrayList<String> result = new ArrayList<String>();
        HashMap<Integer, ArrayList<String>> map = new HashMap<Integer, ArrayList<String>>();

        for (String str : strs) {
            int[] count = new int[26];
            for (int i = 0; i < str.length(); i++) {
                count[str.charAt(i) - 'a']++;
            }

            int hash = getHash(count);
            if (!map.containsKey(hash)) {
                map.put(hash, new ArrayList<String>());
            }

            map.get(hash).add(str);
        }

        for (ArrayList<String> tmp : map.values()) {
            if (tmp.size() > 1) {
                result.addAll(tmp);
            }
        }

        return result;
    }
}

Bootstrap Exercise 1 - Navbar

2016-04-07T11:08:47.000Z

This article is copied from coursera.
https://www.coursera.org/learn/web-frameworks/supplement/BnTE6/exercise-instructions-navbar

Objectives and Outcomes

In this exercise, we will examine the navigation support that we can build into a web page using the Navbar in Bootstrap. We will also learn the basic use of icons in web page design using the built-in glyphicons that are part of Bootstrap, the Font Awesome icons, and bootstrap-social icons. At the end of this exercise, you will be able to:

Create a navigation structure for your website using the Navbar
Use icons within your website to represent various entities making use of the glyphicons, font-awesome icons and bootstrap-social icons
Include additional CSS classes into your project

We will now add a simple navigation bar to the web page so that it provides links to the other pages on the website. Start by adding the following code to the body just above the header jumbotron.

<nav class="navbar navbar-inverse navbar-fixed-top" role="navigation">
    <div class="container">
        <ul class="nav navbar-nav">
            <li class="active"><a href="#">Home</a></li>
            <li><a href="#">About</a></li>
            <li><a href="#">Menu</a></li>
            <li><a href="#">Contact</a></li>
        </ul>
    </div>
</nav>

In the above code, we can see the use of the nav element to specify the navigation information for the website. This nav element is styled by the navbar that declares it as a navigation bar, and the navbar-inverse class to specify that the page should use the dark navigation bar. In addition the inner ul is used to specify the items to be put in the navigation bar. This ul is styled with nav and navbar-nav classes to specify that the items should be displayed inline inside the navigation bar. We also use the container class inside the navigation bar. This navigation bar does not use responsive design at the moment.

We would like the navigation bar elements to collapse for shorter screens, to be replaced by a toggle button so that the items can be toggled on or off when required on smaller screens. This can be achieved by adding the following code to the navigation bar, just below the container div

<div class="navbar-header">
    <button type="button" class="navbar-toggle collapsed" data-toggle="collapse" data-target="#navbar" aria-expanded="false" aria-controls="navbar">
        <span class="sr-only">Toggle navigation</span>
        <span class="icon-bar"></span>
        <span class="icon-bar"></span>
        <span class="icon-bar"></span>
    </button>
    <a class="navbar-brand" href="#">Ristorante Con Fusion</a>
</div>

You will now notice the addition of a link to the left of the Home link with the name of the restaurant. This is the brand name for the website. You can replace this with the logo for the website. This is created by the <a class="navbar-brand"> tag. The other part is the creation of a button with three horizontal lines. For larger screens, this button is hidden. For smaller screens, this button becomes visible. This button will act as the toggle for the navbar items on small screens. At the moment you still see the items being displayed in the navigation bar even for smaller screens. We will fix this in the next step.

To hide the items from the navigation bar for smaller screens, we need to enclose the ul within another navigation bar as follows:

1
2
3

<div id="navbar" class="navbar-collapse collapse">
    <ul class="nav navbar-nav"> ... </ul>
</div>

By doing this, we are specifying that this navbar with the id navbar will be collapsed on smaller screens, but can be toggled on or off when the toggle button is clicked. Note the use of data-toggle="collapse" data-target="#navbar" aria-expanded="false" aria-controls="navbar" within the button. This specifies that the menu items are collapsed on smaller screens when the toggle button is visible. They can be displayed or hidden by clicking the toggle button.

For Chinese Developers: You may find the button can not trigger the navbar list. The reason is that:

1	<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>

and google has been blocked in China!!! You’d better download jquery from website can reference it from local documents. For example:

1	<script src="component/jquery.min.js"></script>

Note that the navbar scrolls when the web page in the browser is scrolled. If we wish to keep the navigation bar always visible at the top of the page when the page is scrolled, then we should change the navbar-static-top to navbar-fixed-top. Let us do this now. Note that after the change, the navigation bar remains visible at the top of the page even when the page is scrolled.
Adding a Dropdown Menu to the Navigation Bar

The next modification adds a Dropdown menu to the navigation bar. Let us target the “Menu” item in the navigation bar and turn it into a dropdown menu item. When this item is clicked, a dropdown menu will be displayed. To do this, modify the list item for the “Menu” item in the navigation bar by replacing it with the following code:

<li class="dropdown">
    <a href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="false"> Menu <span class="caret"></span></a>
    <ul class="dropdown-menu">
        <li><a href="#">Appetizers</a></li>
        <li><a href="#">Main Courses</a></li>
        <li><a href="#">Desserts</a></li>
        <li><a href="#">Drinks</a></li>
        <li role="separator" class="divider"></li>
        <li class="dropdown-header">Specials</li>
        <li><a href="#">Lunch Buffet</a></li>
        <li><a href="#">Weekend Brunch</a></li>
    </ul>
</li>

In this modification, the “Menu” list item is now replaced with a dropdown-toggle, which when pressed will show the selection menu below the navigation bar. The structure of this code has similarities with the toggle button that we used above for showing and hiding the navbar items for small screens.
Modifications to the CSS styles

We would like to have the navigation bar displayed in darker blue color, instead of black. In addition, we would like the navigation bar to indicate the current page by highlighting the item with a darker background in the navbar. In addition, when we use the fixed navigation bar, we should give the body of the page an upper margin of 50px, so that the top 50px of the page does not get hidden under the navigation bar. We accomplish these by adding these CSS customisations to the mystyles.css file

body{
    padding:50px 0px 0px 0px;
    z-index:0;
}

.navbar-inverse {
     background: #303F9F;
}

.navbar-inverse .navbar-nav > .active > a,
.navbar-inverse .navbar-nav > .active > a:hover,
.navbar-inverse .navbar-nav > .active > a:focus {
    color: #fff;
    background: #1A237E;
}

.navbar-inverse .navbar-nav > .open > a,
 .navbar-inverse .navbar-nav > .open > a:hover,
 .navbar-inverse .navbar-nav > .open > a:focus {
    color: #fff;
    background: #1A237E;
}

.navbar-inverse .navbar-nav .open .dropdown-menu> li> a,
.navbar-inverse .navbar-nav .open .dropdown-menu {
    background-color: #303F9F;
    color:#eeeeee;
}

.navbar-inverse .navbar-nav .open .dropdown-menu> li> a:hover {
    color:#000000;
}

We are already beginning to see the page format close to the final format for this module.
Using Icon Fonts and Other CSS classes

The last part of the exercise is to make use of the glyphicons that are provided as part of Bootstrap. In addition we will also use two additional popular font icons.
One of the most popular iconic font toolkit is Font Awesome. Go to its website https://fortawesome.github.io/Font-Awesome/ and download the zip file and move it to the conFusion folder and unzip it. You will find the font-awesome-4.4.0 folder being created. Go into this folder and move the contents of the css folder to conFusion/css and the contents of the fonts folder to conFusion/fonts folder. Then you can delete the font-awesome-4.4.0 folder.
Download the bootstrap-social.css file to conFusion/css folder. This is a modified version of the bootstrap-social available on the http://lipis.github.io/bootstrap-social/. We added in support for G+ and YouTube buttons.
We now need to include the css files for font awesome and bootstrap-social in the index.html file. Add the following code to the head of the file after the links for importing Bootstrap CSS classes:

1 2	<link href="css/font-awesome.min.css" rel="stylesheet"> <link href="css/bootstrap-social.css" rel="stylesheet">

Let us now use some font icons in our web page and decorate it. First use the glyphicons that is part of Bootstrap to add a home icon to the Home link on the navigation bar. Update the home list item as follows:

1 2	<li class="active"><a href="#"><span class="glyphicon glyphicon-home" aria-hidden="true"></span> Home</a></li>

Next, go down to the address in the footer of the page and replace the “Tel.”, “Fax” and “Email” with the corresponding font awesome based icons as follows:

<i class="fa fa-phone"></i>: +852 1234 5678<br>
<i class="fa fa-fax"></i>: +852 8765 4321<br>
<i class="fa fa-envelope"></i>: 
        <a href="mailto:confusion@food.net">confusion@food.net</a>

Finally, let us use the bootstrap-social CSS classes to create the social buttons in the footer by replacing the social sites’ links with the following code:

<div class="nav navbar-nav" style="padding: 40px 10px;">
    <a class="btn btn-social-icon btn-google-plus" href="http://google.com/+"><i class="fa fa-google-plus"></i></a>
    <a class="btn btn-social-icon btn-facebook" href="http://www.facebook.com/profile.php?id="><i class="fa fa-facebook"></i></a>
    <a class="btn btn-social-icon btn-linkedin" href="http://www.linkedin.com/in/"><i class="fa fa-linkedin"></i></a>
    <a class="btn btn-social-icon btn-twitter" href="http://twitter.com/"><i class="fa fa-twitter"></i></a>
    <a class="btn btn-social-icon btn-youtube" href="http://youtube.com/"><i class="fa fa-youtube"></i></a>
    <a class="btn btn-social-icon" href="mailto:"><i class="fa fa-envelope-o"></i></a>
</div>

Conclusions

We have learnt how to add navigation support into a web page using the navigation bar. We also learnt about using other CSS class and other icon fonts in a web project.

LeetCode Solution - Longest Substring Without Repeating Characters

2016-04-03T15:26:16.000Z

https://leetcode.com/problems/longest-substring-without-repeating-characters/

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for “abcabcbb” is “abc”, which the length is 3. For “bbbbb” the longest substring is “b”, with the length of 1.

Solution 1: Using HashSet

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        Set<Character> set = new HashSet<Character>();
        int max = 0, i = 0, j = 0;
        while(j < s.length()){
            if(!set.contains(s.charAt(j))){
                set.add(s.charAt(j++));
                max = Math.max(max, set.size());
            } else {
                set.remove(s.charAt(i++));
            }
        }
        return max;
    }
}

Solution 2: Using ASCII Code

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        boolean[] isUsed = new boolean[256];
        int max = 0, i = 0, j = 0;
        while(j < s.length()){
            if(!isUsed[s.charAt(j)]){
                max = Math.max(max, j - i + 1);
                isUsed[s.charAt(j++)] = true;
            } else {
                isUsed[s.charAt(i++)] = false;
            }
        }
        return max;
    }
}

Follow Up: If you want to return the longest substring not the length of it

public class Solution {
    public String lengthOfLongestSubstring(String s) {
        if(s == null || s.length() == 0){
            return null;
        }
        Set<Character> set = new HashSet<Character>();
        int max = 0, i = 0, j = 0, pos = 0;
        while(j < s.length()){
            if(!set.contains(s.charAt(j))){
                set.add(s.charAt(j++));
                if (max < set.size()) {
                    max = set.size();
                    pos = j - 1;
                }
            }
            else{
                set.remove(s.charAt(i++));
            }
        }
        return s.substring(pos + 1 - max, pos + 1);
    }
}

LeetCode Solution - Longest Palindromic Substring

2016-04-03T02:34:16.000Z

https://leetcode.com/problems/longest-palindromic-substring/

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

Given the string = "abcdzdcab", return "cdzdc".

Solution Reference: http://articles.leetcode.com/longest-palindromic-substring-part-i

Solution 1: Brute Force

// Time Complexity O(N^3)
public class Solution {
    /**
    * @param s input string
    * @return the longest palindromic substring
    */
    public String longestPalindrome(String s) {
        // Write your code here
        for (int i = s.length() - 1; i >= 0; i--) {
            for (int j = 0; j + i < s.length(); j++) {
                if (isPalindrome(s, j, j + i)) {
                    return s.substring(j, j + i + 1);
                }
            }
        }
        return null;
    }
    private boolean isPalindrome(String s, int start, int end) {
        for (int i = start, j = end; i < j; i++, j--) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
        }
        return true;
    }
}

Solution 2: Dynamic Programming

1	Define P[ i, j ] ← true iff the substring Si … Sj is a palindrome, otherwise false.

Therefore,

1	P[ i, j ] ← ( P[ i+1, j-1 ] and Si = Sj )

The base cases are:

1 2	P[ i, i ] ← true P[ i, i+1 ] ← ( Si = Si+1 )

// Time Complexity O(N^2), Space Complexity O(N^2)
public class Solution {
    /**
     * @param s input string
     * @return the longest palindromic substring
     */
    public String longestPalindrome(String s) {
        // Write your code here
        boolean[][] isPalindrome = new boolean[s.length()][s.length()];
        isPalindrome[s.length() - 1][s.length() - 1] = true;
        int start = 0, maxLength = 1;
        for (int i = 0; i < s.length() - 1; i++) {
            isPalindrome[i][i] = true;
            if (s.charAt(i) == s.charAt(i + 1)) {
                isPalindrome[i][i + 1] = true;
                start = i;
                maxLength = 2;
            }
        }
        for (int pos = 2; pos < s.length(); pos++) {
            for (int i = 0; i < s.length() - pos; i++) {
                int j = i + pos;
                if (isPalindrome[i + 1][j - 1] && s.charAt(i) == s.charAt(j)) {
                    isPalindrome[i][j] = true;
                    start = i;
                    maxLength = pos + 1;
                }
            }
        }        
        return s.substring(start, start + maxLength);
    }
}

Solution 3

// Time Complexity: O(N^2), Space Complexity O(1)
public class Solution {
    /**
     * @param s input string
     * @return the longest palindromic substring
     */
    public String longestPalindrome(String s) {
        // Write your code here
        String result = "";
        if (s == null) {
            return result;
        }
        String result1 = "";
        String result2 = "";
        int maxLength = 0;
        for (int i = 0; i < s.length(); i++) {
            result1 = expandPalindrome(s, i, i);
            if (result1.length() > maxLength) {
                maxLength = result1.length();
                result = result1;
            }
            result2 = expandPalindrome(s, i, i + 1);
            if ( result2.length() > maxLength ){
                maxLength = result2.length();
                result = result2;
            }
        }
        return result;
    }
    private String expandPalindrome(String s, int start, int end) {
        while (start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
            start--;
            end++;
        }
        return s.substring(start + 1, end);
    }
}

LeetCode Solution - Letter Combinations of a Phone Number

2016-03-30T06:18:08.000Z

https://leetcode.com/problems/letter-combinations-of-a-phone-number/

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string “23”
Output: [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

This is problem is easy except the use of HashMap.

Java has three kinds of Hashtalbe implementations. They are HashTable, HashMap, HashSet. HashTable is a thread safe implementation, but it is not efficient. For a single thread problem, we always use HashMap.

HashMap have several important methods. They are put(key, value), get(key), keySet(), values().

`put(key, value)`

put(key value) will insert the pair key-value into hashtable. For example:

Map<String,String> map = new HashMap<String,String>();
map.put("one","xiaoming");      // it doesn't have add method since 
map.put("two","xiaozhang");     // it does not implement Collection 
map.put("three","xiaoli");      // interface
map.put("four","xiaoliu");

one, two, three, four are keys and xiaoming, xiaozhang, xiaoli, xiaoliu are values.

`get(key)`

get(key) will return value which combines with key. For example:

System.out.println(map.get("one"));
System.out.println(map.get("two"));
System.out.println(map.get("three"));
System.out.println(map.get("four"));

The output will be:

xiaoming
xiaozhang
xiaoli
xiaoliu

`keySet()`

keySet() will return all keys stored in hashtable.

System.out.println(map.keySet());

The output is:

[four, one, two, three]

`values()`

values() will return all values stored in hashtable.

System.out.println(map.value());

The output is:

[xiaolu, xiaoming, xiaozhang, xiaoli]

Solution


/**
 * 本代码由九章算法编辑提供。没有版权欢迎转发。
 * - 九章算法致力于帮助更多中国人找到好的工作，教师团队均来自硅谷和国内的一线大公司在职工程师。
 * - 现有的面试培训课程包括：九章算法班，系统设计班，BAT国内班
 * - 更多详情请见官方网站：http://www.jiuzhang.com/
 */
 
public class Solution {
    public ArrayList<String> letterCombinations(String digits) {
        ArrayList<String> result = new ArrayList<String>();

        if (digits == null || digits.equals("")) {
            return result;
        }

        Map<Character, char[]> map = new HashMap<Character, char[]>();
        map.put('0', new char[] {});
        map.put('1', new char[] {});
        map.put('2', new char[] { 'a', 'b', 'c' });
        map.put('3', new char[] { 'd', 'e', 'f' });
        map.put('4', new char[] { 'g', 'h', 'i' });
        map.put('5', new char[] { 'j', 'k', 'l' });
        map.put('6', new char[] { 'm', 'n', 'o' });
        map.put('7', new char[] { 'p', 'q', 'r', 's' });
        map.put('8', new char[] { 't', 'u', 'v'});
        map.put('9', new char[] { 'w', 'x', 'y', 'z' });

        StringBuilder sb = new StringBuilder();
        helper(map, digits, sb, result);

        return result;
    }

    private void helper(Map<Character, char[]> map, String digits, 
        StringBuilder sb, ArrayList<String> result) {
        if (sb.length() == digits.length()) {
            result.add(sb.toString());
            return;
        }

        for (char c : map.get(digits.charAt(sb.length()))) {
            sb.append(c);
            helper(map, digits, sb, result);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}

LeetCode Solution - Generate Parentheses

2016-03-30T04:17:23.000Z

https://leetcode.com/problems/generate-parentheses/

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

This is an easy problem. The keyword is all combinations.


public class Solution {
    /**
    * @param n n pairs
    * @return All combinations of well-formed parentheses
    */
    public ArrayList<String> generateParenthesis(int n) {
        // Write your code here
        ArrayList<String> result = new ArrayList<String>();
        String temp = new String();

        helper(result, temp, n, 0, 0);

        return result;
    }
    private void helper(ArrayList<String> result, String temp, int n,
                            int leftParentheses, int rightParentheses) {
        if (temp.length() == 2 * n) {
            result.add(temp);
        }
        
        if (leftParentheses < n) {
            helper(result, temp + '(', n, leftParentheses + 1, rightParentheses);
        }
        if (rightParentheses < leftParentheses) {
            helper(result, temp + ')', n, leftParentheses, rightParentheses + 1);
        }
        
        return;
    }
}

LeetCode Solution - Restore IP Addresses

2016-03-30T02:52:18.000Z

https://leetcode.com/problems/restore-ip-addresses/

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given "25525511135",
return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

Keywords: ALL POSSIBLE gives us hint to use DFS

Another important point is that we should eliminate the case like '00', '01', '001', etc.


public class Solution {
    /**
    * @param s the IP string
    * @return All possible valid IP addresses
    */
    public ArrayList<String> restoreIpAddresses(String s) {
        // Write your code here
        char[] str = s.toCharArray();
        int pointNums = 0;
        ArrayList<String> result = new ArrayList<String>();
        String temp = new String();
    
        if (s == null || s.length() < 4 || s.length() > 12) {
            return result;
        }
        restoreIpAddressesHelper(result, temp, str, 0, pointNums);
        return result;
    }

    private void restoreIpAddressesHelper(ArrayList<String> result,
                                            String temp,
                                            char[] str,
                                            int pos,
                                            int pointNums) {
        if (temp.length() == str.length + 3 && pointNums == 3) {
            result.add(temp);
        }
        if (pointNums > 3) {
            return;
        }
    
        StringBuffer addString = new StringBuffer();
        for (int i = pos; i < pos + 3 && i < str.length; i++) {
            addString.append(str[i]);
            if (!isValidHelper(addString.toString())) {
                break;
            }
            if ( i == str.length - 1) {
                restoreIpAddressesHelper(result, temp + addString, 
                                            str, i + 1, pointNums);
            } else {
                restoreIpAddressesHelper(result, temp + addString + ".", 
                                            str, i + 1, pointNums + 1);
            }
        }
        return;     
    }
    private boolean isValidHelper(String s) {
        if (s.charAt(0) == '0' && s.length() != 1) {
            return false;
        } else if (Integer.valueOf(s) > 255 ) {
            return false;
        } else {
            return true;
        }
    }
}

LeetCode Solution - Next Permutation

2016-03-29T13:48:40.000Z

https://leetcode.com/problems/next-permutation/

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

The test case given in the question is not enough for us to understand this question. So I will use another example to explain this question.

Let 1,2,3,4,5 is the input array nums. It is easy to write down its next permutation which is 1,2,3,5,4. What about 1,2,3,5,4‘s next permutation? It’s 1,2,4,3,5. And why?

The reason is that we have listed all possible permutations that index[nums.length - 2] greater than index[nums.length - 1], we should think about condition that the last three numbers contain a non-lexicographically permutation of numbers. The smallest one is that index[nums.length - 3] is the second smallest number in the last three numbers, and numbers after that should be in an ascending order. Why numbers should be in an ascending order? Because the question defines that this is a smallest condition.

The following should be 1,2,4,5,3, 1,2,5,3,4, 1,2,5,4,3, 1,3,2,4,5, 1,3,2,5,4, 1,3,4,2,5, 1,3,4,5,2, 1,3,5,2,4, 1,3,5,4,2, 1,4,2,3,5, 1,4,2,5,3, 1,4,3,2,5, 1,4,3,5,2, 1,4,5,2,3, 1,4,5,3,2, etc.


public class Solution {
    public void nextPermutation(int[] nums) {
        if (nums == null || nums.length == 0 || nums.length == 1) {
            return;
        }
        
        int index = -1;
        // find the position where the value is smaller than its follower
        for (int i = nums.length - 1; i > 0; i--) {
            if (nums[i] > nums[i - 1]) {
                index = i - 1;
                break;
            }
        }
        // if index equals -1, this means the array is in an non-ascending,
        // order, it doesn't have next permutation, or it's next permutation
        // is the lowest possible order (ie, sorted in ascending order).  
        if (index == -1) {
            reverse(nums, 0, nums.length - 1);
            return;
        }
        // find the position where the value is greater than the value stored
        // in nums[index]. Remember that index is the position where we need to
        // make some changes.
        for (int i = nums.length - 1; i > index; i--) {
            // exchange 
            // e.g. for 5,4,7,5,3,2; index = 1, i = 3
            // The result will be 5,5,7,4,3,2
            // Numbers after i = 3 (nums[i] = 5) is in an non-ascending order,
            // which means it's the greatest possible order. So the next   
            // permutation should start with the number smallest greater than 5. 
            // And the number after index will also in an non-ascending order.
            if(nums[i] > nums[index]){
                int temp = nums[i];
                nums[i] = nums[index];
                nums[index] = temp;
                break;
            }
        }
        // once we get 5,5,7,4,3,2 from 5,4,7,5,3,2, we should reverse the order 
        // from 7 to 2 to get the lowest possible order.
        reverse(nums, index + 1, nums.length - 1);
        return;
    }
    private void reverse(int[] nums, int start, int end) {
        for (int i = start, j = end; i < j; i++, j--) {
            int temp = nums[j];
            nums[j] = nums[i];
            nums[i] = temp;
        }
    }
}

Another version of solution which helps me a lot to understand this question.

https://leetcode.com/discuss/85045/20-line-1ms-in-place-java-code-with-expalantion

scan from right to left, find first pair where a[i] > a[i-1]. note that a[i:] is non-ascending.
scan from right to left again, find first element a[j] where a[j] > a[i-1]. since a[i:] is non-ascending, a[j] is the smallest number that > a[i-1].
swap a[i-1] with a[j]. note that after swap, a[i:] remains non-ascending.
reverse a[i:] so that a[i:] becomes non-descending.


public class Solution {
    public void nextPermutation(int[] nums) {
        int p = nums.length-1;
        while (p>0 && nums[p]<=nums[p-1]) { --p; }
        if (p == 0) {  // case like [3,2,1]
            reverse(nums, 0, nums.length-1);
            return;
        }
        int q = nums.length-1;
        while (nums[q]<=nums[p-1]) { --q; }
        int temp = nums[p-1]; nums[p-1] = nums[q]; nums[q] = temp;
        reverse(nums, p, nums.length-1);
    }

    private void reverse(int[] a, int from, int to) {
        for (; from < to; ++from, --to) {
            int temp = a[from];
            a[from] = a[to];
            a[to] = temp;
        }
    }
}

LintCode Solution - String Permutation II

2016-03-29T13:48:40.000Z

http://www.lintcode.com/en/problem/string-permutation-ii/

Given a string, find all permutations of it without duplicates.
Example
Given “abb”, return [“abb”, “bab”, “bba”].

Given “aabb”, return [“aabb”, “abab”, “baba”, “bbaa”, “abba”, “baab”].

We can see the keywords in the description of the question is that “find all permutations”. It is easy for us to think that we can use dfs to solve this problem.


public class Solution {
    /**
    * @param str a string
    * @return all permutations
    */
 
    public List<String> stringPermutation2(String str) {
        // Write your code here
        char[] string = str.toCharArray();
        boolean[] isUsed = new boolean[string.length];
        Arrays.sort(string);
        List<String> result = new ArrayList<String>();
        String temp = new String();
        stringPermutation2Helper(result, temp, string, isUsed);
        return result;
    }
    
    private void stringPermutation2Helper(List<String> result, 
                                            String temp,
                                            char[] string,
                                            boolean[] isUsed) {
        if (temp.length() == string.length) {
            result.add(temp);
            return;
        }
        for (int i = 0; i < string.length; i++) {
            if (isUsed[i] == true || i != 0 && 
                isUsed[i - 1] == false && 
                string[i] == string[i - 1]) {
                continue;
            }
            isUsed[i] = true;
            stringPermutation2Helper(result, temp + string[i], string, isUsed);
            // backtracking
            isUsed[i] = false;
        }
    }
}

This is a navie version of solution. It needs extra memory which is O(n), where the n represents the length of the input string, to store the information if this char in string has been added. The time complexity of this algorithm is O(n^2).

If you want to solve this problem without extra memory space, we can use solution used in “Next Permutation”


/**
* 本代码由九章算法编辑提供。没有版权欢迎转发。
* - 九章算法致力于帮助更多中国人找到好的工作，教师团队均来自硅谷和国内的一线大公司在职工程师。
* - 现有的面试培训课程包括：九章算法班，系统设计班，BAT国内班
* - 更多详情请见官方网站：http://www.jiuzhang.com/
*/

public class Solution {

    /**
    * @param str a string
    * @return all permutations
    */
 
    public List<String> stringPermutation2(String str) {
        // Write your code here
        List<String> result = new ArrayList<String>();
        char[] s = str.toCharArray();
        Arrays.sort(s);
        result.add(String.valueOf(s));
        while ((s = nextPermutation(s)) != null) {
            result.add(String.valueOf(s));
        }
        return result;
    }
    
    public char[] nextPermutation(char[] nums) {
        int index = -1;
        for(int i = nums.length -1; i > 0; i--){
            if(nums[i] > nums[i-1]){
                index = i-1;
                break;
            }
        }
        if(index == -1){
            return null;
        }
        for(int i = nums.length -1; i > index; i--){
            if(nums[i] > nums[index]){
                char temp = nums[i];
                nums[i] = nums[index];
                nums[index] = temp;
                break;
            }
        }
        reverse(nums,index+1,nums.length-1);
        return nums;
    }
    
    public void reverse(char[] num, int start, int end) {
        for (int i = start, j = end; i < j; i++, j--) {
            char temp = num[i];
            num[i] = num[j];
            num[j] = temp;
        }
    }
}

CSS 定位

2016-03-20T10:27:07.000Z

这篇笔记是我在上Baidu IFE课程进行Task Three时整理的一些资料。

CSS定位

普通流：元素按照其在HTML中的位置顺序决定分布的过程
浮动
绝对布局

CSS position属性：

普通流

编写html和css代码如下：

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>定位</title>
    <link href="style.css" type="text/css" rel="stylesheet">
</head>
<body>
    <div id="position1"></div>
    <script>
    for (var i = 0; i<100; i++) {
            document.write(i+"<br/>")
        }
    </script>
</body>
</html>

#position1{
    width: 100px;
    height: 100px;
    background-color: aqua;
}

所谓普通流就是指按照html代码里从上到下的顺序布局区块。上面代码的显示效果如下：

绝对布局`absolute`

通过添加style.css文件中position，我们可以改变其布局方式。当position的属性值为absolute时，其相当于不占据浏览器位置，可在其所占位置继续填充内容。代码如下：

#position1{
    width: 100px;
    height: 100px;
    background-color: aqua;
    position: absolute;
}

在浏览器中显示效果如下：

`fixed`属性：

fixed效果如下：

`relative`属性：

relative中的top,right,bottom,left是相对于它不添加属性的起始位置所占据的空间而言的。

`z-index`属性

原本存在这样一种情况，两个relative的方块有部分重合，且红色的位于蓝色之上：

当我们分别在两个position中添加z-index的值之后，我们可以改变两个区块的位置。两个重叠的区块，谁的的z-index的值大，谁就呈现在前面。注意：z-index的值是没有上限的，而且同样适用于绝对布局absolute

CSS float浮动：

float属性可用的值：

left: 元素向左浮动
right: 元素向右浮动
none: 元素不浮动
inherit: 从父级继承浮动属性

原本浏览器中存在着如下的布局：

当我们在#fd中添加float: left时，我们可以看到蓝色的正方形块消失了。其实，是蓝色区块被红色区块给挡住了。与绝对布局相似，红色区块不在占用浏览器区域。

当三个区块都设置float: left属性性时，我们可以发现，三个区块并排在顶部左侧，但不会重叠。

当我们在三个左浮动的<div>的外侧在添加一个container，同时这个container的width=200px小于其内部三个<div>的width之后。我们可以看到，绿色的区块换行了。这是因为主容器无法承载了。如果这时红色的height=200px，那么绿色区块出现的位置是在蓝色区块的正下方而非红色区块正下方。

CSS 定位-浮动应用：瀑布流

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>定位</title>
    <link href="style.css" type="text/css" rel="stylesheet">
</head>
<body>
<div id="div1">
      <ul>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F1.jpg"></li>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F2.jpg"></li>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F3.jpg"></li>
      </ul>
      <ul>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F4.jpg"></li>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F5.jpg"></li>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F6.jpg"></li>
      </ul>
      <ul>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F7.jpg"></li>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F8.jpg"></li>
      <li><img src="http://7xs1zn.com1.z0.glb.clouddn.com/img%2Fcss%2F9.jpg"></li>
      </ul>
</div>
</body>
</html>

* {
    margin: 0;
    padding: 0;
}
li {
    list-style: none;
}
#div1 {
    width: 950px;
    height: auto;
    margin: 20px auto;
}
ul {
    width: 250px;
    float: left;
}

上面代码的执行效果如下：

LeetCode Solution - Decode Ways

2016-03-19T15:27:07.000Z

https://leetcode.com/problems/decode-ways/

A message containing letters from A-Z is being encoded to numbers using the following mapping:

‘A’ -> 1
‘B’ -> 2
…
‘Z’ -> 26

Given an encoded message containing digits, determine the total number of ways to decode it.

EXAMPLE: Given encoded message 12, it could be decoded as AB (1 2) or L (12).
The number of ways decoding 12 is 2.

First, we need a array with length eqauls s.length() + 1, since we need to add 0 before the given string just for convience. (p.s. we will do similar thing for DP problem in many cases)

Let array nums store the value in string, array result store the decode ways for string.

There are two kinds of conditions:

the value in nums[i] that you are dealing with is not 0
the value in nums[i] that you are dealing with eqauls 0

For the first condition, since nums[i] is not 0, it can convert to a meaningful charater by itself. So the total possible decode ways, result[i], at least same as result[i - 1].
Then we consider if it is able to combine with nums[i - 1]. If the value nums[i - 1] * 10 + nums[i] is greater than 10 and smaller than 26, which means nums[i - 1] * 10 + nums[i] is a available value, result[i] is able to add result[i - 1]. Otherwise, it is not able to do that.

For the second condition, since nums[i] is 0, it can not convert to a meaningful character by itself. It must combine with nums[i - 1]. So we could assume nums[i - 1] * 10 + nums[i] is the value we are dealing with. the result[i] can only possibly come from result[i - 2]. The following operation is same as condition 1.

public class Solution {
    public int numDecodings(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        int[] nums = new int[s.length() + 1];
        // 在string前面加了一个0
        nums[0] = 1;
        nums[1] = s.charAt(0) != '0' ? 1 : 0;
        for (int i = 2; i <= s.length(); i++) {
            if (s.charAt(i - 1) != '0') {
                nums[i] = nums[i - 1];
            }
            int twoDigits = (s.charAt(i - 2) - '0') * 10 + s.charAt(i - 1) - '0';
            if (twoDigits >= 10 && twoDigits <= 26) {
                nums[i] += nums[i - 2];
            }
        }
        return nums[s.length()];
    }
}

CSS Box Model

2016-03-19T15:27:07.000Z

这篇笔记是我在上Baidu IFE课程进行Task Two时整理的一些资料。

CSS盒子模型

reference: http://www.runoob.com/css/css-tutorial.html and http://www.jikexueyuan.com/course/164_6.html?ss=1

CSS盒子模型概述

CSS盒子模型中，内部红色部分为内容，绿色部分为内边界，黄色线处为边框，外部银白色处外边距。
可以看到height与width所指定的高和宽为内容部分的尺寸。

内边距

内边距在content外，边框内

内边距属性：

示例：

1	<td style="padding:100px">内边距</td>

上面代码的执行效果如下，会在表格内部<td> 上下左右各填充100px的空余空间。padding-right , padding-left , padding-top , padding-bottom 具有类似的语义。

边框

CSS边框：
- 可以创建出效果出色的边框，并且可以应用于任何元素
边框样式：
- border-style: 定义了10个不同的非继承样式，包括none
边框的单边样式：
- border-top-style
- border-left-style
- border-right-style
- border-bottom-style
边框的宽度：
- border-width
边框的单边宽度：
- border-top-width
- border-left-width
- border-right-width
- border-bottom-width
边框的颜色：
- border-color
边框的单边颜色：
- border-top-color
- border-left-color
- border-right-color
- border-bottom-color

`border-style` 测试示例：

我们在style.css 中编写如下的代码：

p{
    width: 100px;
    border-top-style: double;
    border-left-style: dotted;
    border-right-style: groove;
    border-bottom-style: solid;
}

在我们的html编写如下代码:

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Title</title>
    <link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
    <p>CSS边框测试</p>
</body>
</html>

在网页中显示效果如下：

`border-width` 测试示例：

我们在style.css 中添加代码成如下：

p{
    width: 100px;
    border-top-style: double;
    border-left-style: dotted;
    border-right-style: groove;
    border-bottom-style: solid;
    border-bottom-width: 10px;
}

在网页中显示效果如下：

其他一些测试示例：

p{
    width: 200px;
    border-radius: 10px;
    background-color: aqua;
    text-align: center;
    border: 2px solid red;
    box-shadow: 10px 10px 5px chartreuse;
}

效果如下：

其中：

border-radius 用于设置边框的圆角；
border 用于设计边框的边界，第一个参数是线的粗细，第二个参数为边框边界的样式，第三个为颜色；
box-shadow 用于设计边框阴影：第一个参数表示向右移动10px ，第二个参数表示向下移动10px ，第三个参数表示阴影的透明度，第四个参数表示阴影的颜色。

外边距

Margin

margin清除周围的元素（外边框）的区域。margin没有背景颜色，是完全透明的
margin可以单独改变元素的上，下，左，右边距。也可以一次改变所有的属性。

如下图所示，css代码中为设计body 的margin 时，html代码中<div> 所包裹的部分无法覆盖整个浏览器。这就是body的外边距。

当在css中添加如下代码时：

1
2
3

body{
margin: 0px;
}

显示效果如下：

盒子模型设计综述

盒子模型在html中用<div> 来表示，如下：

<div class="container">
	<div class="bd">
		<div class="pd">
			<div class="content">
				CSS盒子模型
			</div>
		</div>
	</div>
</div>

在css用代码如下：

body{
    margin: 0px;
}
.container{
    margin: 10px;
}
.bd{
    border-style: dotted;
    border-color: #ffccff;
}
.pd{
    padding:10px;
}
.content{
    background-color: green;
    text-align: center;
    color: red;
}

最后显示效果如下：

外边距合并

外边距合并就是一个叠加的概念

两个盒子会自动合并。

<div class="container">
	<div class="bd">
		<div class="pd">
			<div class="content">
				CSS盒子模型
			</div>
		</div>
	</div>
</div>
<div class="container2">
	<div class="bd2">
		<div class="pd2">
			<div class="content2">
				CSS盒子模型2
			</div>
		</div>
	</div>
</div>

body{
    margin: 0px;
}
.container{
     margin: 100px;
 }
.bd{
    border-style: dotted;
    border-color: #ffccff;
}
.pd{
    padding:10px;
}
.content{
    background-color: green;
    text-align: center;
    color: red;
}
.container2{
    margin: 100px;
}
.bd2{
    border-style: dotted;
    border-color: #ffccff;
}
.pd2{
    padding:10px;
}
.content2{
    background-color: green;
    text-align: center;
    color: red;
}

显示效果如下：

如果改变.container{} 中margin 的值为50px ，则显示效果如下：

merge合并所遵循是多的那一部分。

### CSS3中盒子的类型

<div> 属于block，<span> 属于inline，他们的显示效果如下，其中<div>占据两行属性，<span>为内联元素，占据一行。

通过在<style>中定义display可以改变其显示属性，代码如下：

<style>
	div{
		background-color: aqua;
		display: inline;
	}
	span{
		background-color: greenyellow;
		display: block;
	}
</style>

显示效果如下：

`inline-block`属性

inline-block与inline具有相同的显示类型，那么为什么要出现inline-block类型呢？
inline 不能改变宽度和高度，但inline-block可以设置宽度和高度。如下，我们设置inline-block和inline的width属性都为300px，代码如下：

<style>
	div{
		background-color: aqua;
		display: inline-block;
		width: 300px;
	}
	span{
		background-color: greenyellow;
		display: inline;
		width: 300px;
	}
</style>

显示效果如下：

Hello World

2016-03-19T14:42:28.000Z

Welcome to Hexo! This is your very first post. Check documentation for more info. If you get any problems when using Hexo, you can find the answer in troubleshooting or you can ask me on GitHub.

Quick Start

Create a new post

1	$ hexo new "My New Post"

More info: Writing

Run server

1	$ hexo server

More info: Server

Generate static files

1	$ hexo generate

More info: Generating

Deploy to remote sites

1	$ hexo deploy

More info: Deployment

Entry-Level段子手的日常

Missing Number

Solution 1: Binary Search

Solution 2: Extra O(N) Space

Solution 3: Bit Manipulation

Solution 4: Method Without Bit Manipulation

Bit Manipulation

Convert a decimal number into binary number with bit manipulation

LeetCode Example 1 - Number of 1 Bits

Solution 1

Solution 2 O(logN)

Solution 3 O(countOnes(N))

LeetCode Example 2 - Reverse Bits

LeetCode Solution - Single Number

Single Number I

Solution 1: With HashSet

Solution 2: With Sort Algorithm

Solution 3: With XOR Operation

Single Number II

Solution 1: With HashMap

Solution 2: With Bit Manipulation

Single Number III

Solution 1: With HashMap

Solution 2: With XOR Operation

More Reference:

LintCode Solution - Binary Representation

Solution

ATTENTION

LintCode Solution - Anagrams

Solution 1

Solution 2

Bootstrap Exercise 1 - Navbar

Exercise: Navbar

Objectives and Outcomes

Create a basic navigation bar

Creating a responsive navigation bar

Conclusions

LeetCode Solution - Longest Substring Without Repeating Characters

Solution 1: Using HashSet

Solution 2: Using ASCII Code

Follow Up: If you want to return the longest substring not the length of it

LeetCode Solution - Longest Palindromic Substring

Solution 1: Brute Force

Solution 2: Dynamic Programming

Solution 3

LeetCode Solution - Letter Combinations of a Phone Number

put(key, value)

get(key)

keySet()

values()

Solution

LeetCode Solution - Generate Parentheses

LeetCode Solution - Restore IP Addresses

LeetCode Solution - Next Permutation

LintCode Solution - String Permutation II

CSS 定位

CSS定位

CSS position属性：

普通流

绝对布局absolute

fixed属性：

relative属性：

z-index属性

CSS float浮动：

CSS 定位-浮动应用：瀑布流

LeetCode Solution - Decode Ways

CSS Box Model

CSS盒子模型

CSS盒子模型概述

内边距

内边距属性：

边框

border-style 测试示例：

border-width 测试示例：

其他一些测试示例：

外边距

Margin

盒子模型设计综述

外边距合并

inline-block属性

`put(key, value)`

`get(key)`

`keySet()`

`values()`

绝对布局`absolute`

`fixed`属性：

`relative`属性：

`z-index`属性

`border-style` 测试示例：

`border-width` 测试示例：

`inline-block`属性