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 

 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:

Input
 
0.5

Output
 
.1

Expected
 
0.1



Input
 
1.0

Output
 
1.

Expected
 
1

 

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