LeetCode--Missing Number

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?

代码:

    int missingNumber(vector<int>& nums) {        int x = 0;        int y = 0;        int z = 0;        for(int i = 0; i < nums.size()+1; ++i)        {            x ^= i;        }        for(int i = 0; i < nums.size(); ++i)        {            y ^= nums[i];        }        z = x^y;        return z;    }

运用异或,x为0~n异或,y为nums所有值异或,x^y即剩下miss的那个