268. Missing Number

This idea comes from the problem of "136. Single Number", i.e. xor cancels the same number. So we can xor all the numbers from 0 to n first. And then xor the numbers in the input array. The remaining number must be the missing number.

1:  class Solution {  
2:  public:  
3:    int missingNumber(vector<int>& nums) {  
4:      int n = nums.size();  
5:      int res = 0;  
6:      for (int i = 0; i <= n; i++) {  
7:        res ^= i;  
8:      }  
9:      for (int i = 0; i < nums.size(); i++) {  
10:        res ^= nums[i];  
11:      }  
12:      return res;  
13:    }  
14:  };