137. Single Number II

3/20/2018 update:
这道题另外一种解法真是太难了。看不懂，性价比超低。

Analysis:
Use array [14,14,14,9] as an example. Translate every number to binary, we can get:

1110

1110

1110

1001

-------

4331 and sum up each position, we can get the result binary from %3 on each position.

Solution:

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