268. Missing Number

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?

一刷
类似于寻找哪个数字只出现了一次，其它的都出现两次。
如果[0, 2, .., n-1]中间有一个数字miss, 数组长度n-1
那么每次与数组中的元素异或，与index异或，最后留下的数字为miss

public class Solution {
    public int missingNumber(int[] nums) {
        int res = 0;
        for(int i=0; i<nums.length; i++){
            res ^= i^nums[i];
        }
        res ^= nums.length;
        return res;
    }
}