137. Single Number II

 1 class Solution {
 2 public:
 3     int singleNumber(vector<int>& nums) {
 4         int len = sizeof(nums[0]) * 8;
 5         vector<int> count(len, 0);
 6         for (size_t i = 0; i < nums.size(); ++i) {
 7             int cmp = 1;
 8             for (int j = 0; j < len; ++j) {
 9                 //if (nums[i] < cmp) break; // 要特别注意这里，数值可能为负值，这样判小就会出现错误
10                 if (nums[i] & cmp) {
11                     ++count[j];
12                     count[j] = count[j] % 3;
13                 }
14                 cmp <<= 1;
15             }
16         }
17         int num = 0;
18         for (int i = 0; i < len; ++i) {
19             if (count[i]) num += (count[i] << i);// 如果调用pow函数，如果是负值就会出现错误，要考虑全面
20         }
21         return num;
22     }
23 };