力扣算法——137SingleNumberII【M】

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

Solution：
　　因为不能使用额外的空间，不然就可以开辟数组来对数据个数进行统计了
　　那么只能使用位运算了，计算32位中，每个数字出现1的次数，一旦个数不为1，那么该位就是那个唯一数字的位数了

 class Solution {
 public:
     int singleNumber(vector<int>& nums) {
         int bits[] = {  };
         for (int i = ; i < nums.size(); ++i)
         {
             int c = ;
             for (int j = ; j < ; ++j)
             {
                 bits[j] += (c & nums[i]) ?  : ;
                 c = c << ;
             }
         }
         int res = ;
         for (int i = ; i < ; ++i)
             if (bits[i] % )
                 res += (int)pow(2.0, i);
         return res;
     }
 };