题目：

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,

Given nums = [0, 1, 3] return 2.

Note:

Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

解析：

1 class Solution {2 public:3     int missingNumber(vector
     
      & nums) {4         sort(nums.begin(),nums.end());5         for(int i = 0; i <= nums.size(); i++)6             if(nums[i] != i)7                 return i;8     }9 };
     

高效的算法应该是使用位运算，我没明白，也不想深究：

class Solution {public:    int missingNumber(vector
     
      & nums) {        int result = 0;        for (int i = 0; i < nums.size(); i++)            result ^= nums[i]^(i+1);        return result;    }};