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

重点在于数组是从0开始的;

The basic idea is to use XOR operation. We all know that a^b^b =a, which means two xor operations with the same number will eliminate the number and reveal the original number.
In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing number.

例如：

index:0， 1， 2， ...， k-1， k， ...， n-1

elements： 0， 1， 2， ..., k-1， k+1，  ...， n

最后分别剩下了k和n，最后再与n异或，就只剩下k了。骚

参考：https://leetcode.com/problems/missing-number/discuss/69791/4-Line-Simple-Java-Bit-Manipulate-Solution-with-Explaination