leetcode 268 Missing Number(异或运算的应用)

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?

题解：求0到n的和s,再求数组的和s2，s-s2就是少的那个数。

由于计算机中异或运算比加法运算快。而且这道题正好可以用异或的性质来解决。

两个相同的数异或结果是0.

0和其他数异或结果还是那个数。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int n=nums.size();
        int s=0;
        for(int i=1;i<=n;i++){
            s^=i;
        }
        int s2=nums[0];
        for(int i=1;i<n;i++){
            s2^=nums[i];
        }
        return s^s2;
    }
};