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LintCode Solution - Binary Representation

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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

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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('.')));