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”Examplen = 3.72, return ERRORn = 3.5, return 11.1

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

1. use%2to 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

 1 public class Solution { 2     /** 3      *@param n: Given a decimal number that is passed in as a string 4      *@return: A string 5      */ 6     public String binaryRepresentation(String n) { 7         int intPart = Integer.parseInt(n.substring(0, n.indexOf('.'))); 8         double decPart = Double.parseDouble(n.substring(n.indexOf('.'))); 9         String intstr = "";10         String decstr = "";11         12         if (intPart == 0) intstr += '0';13         while (intPart > 0) {14             int c = intPart % 2;15             intstr = c + intstr;16             intPart = intPart / 2;17         }18        19         while (decPart > 0.0) {20             if (decstr.length() > 32) return "ERROR";21             double r = decPart * 2;22             if (r >= 1.0) {23                 decstr += '1';24                 decPart = r - 1.0;25             }26             else {27                 decstr += '0';28                 decPart = r;29             }30         }31         return decstr.length() > 0? intstr + "." + decstr : intstr;32     }33 }

这道题还有一些细节要处理，我之前忽略了

比如：

0.5

.1

0.1

1.0

1.

1