Lintcode: 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 number can not be represented accurately in binary, print “ERROR”

Example
n = 3.72, return ERROR

n = 3.5, return 11.1

For int part, similar approach of extracting numbers from int:

1. use %2 to get each digit from lowest bit to highest bit.

2. int right shift 1 position (=>>1).

3. construct the binary number (always add to the higher position of the current binary number)

Please refer to the code below for the process above.

For decimal part, use *2 approach.  For example:

int n = 0.75

n*2 = 1.5

Therefore, the first digit of binary number after '.' is 1 (i.e. 0.1).  After constructed the first digit, n= n*2-1 

public class Solution {
    /**
     *@param n: Given a decimal number that is passed in as a string
     *@return: A string
     */
    public String binaryRepresentation(String n) {
        int intPart = Integer.parseInt(n.substring(0, n.indexOf('.')));
        double decPart = Double.parseDouble(n.substring(n.indexOf('.')));
        String intstr = "";
        String decstr = "";
        
        if (intPart == 0) intstr += '0';
        while (intPart > 0) {
            int c = intPart % 2;
            intstr = c + intstr;
            intPart = intPart / 2;
        }
       
        while (decPart > 0.0) {
            if (decstr.length() > 32) return "ERROR";
            double r = decPart * 2;
            if (r >= 1.0) {
                decstr += '1';
                decPart = r - 1.0;
            }
            else {
                decstr += '0';
                decPart = r;
            }
        }
        return decstr.length() > 0? intstr + "." + decstr : intstr;
    }
}

这道题还有一些细节要处理，我之前忽略了

比如：

Input

 

0.5

Output

 

.1

Expected

 

0.1

Input

 

1.0

Output

 

1.

Expected

 

1