[LeetCode]Missing Number

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

这题需要技巧，利用异或。

先将nums的数作异或得到x1；

将0-n+1都作异或得到x2；

missing number 就是x1^x2。

为什么呢？

假设缺少的值为A，则x2=x1^A，那么x1^x2=x1^(x1^A)=(x1^x1)^A=0^A=A，得证。

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int result;
        int x1=nums[0];
        for(int i=1;i<nums.size();i++)
        {
            x1^=nums[i];
        }
        int x2=0;
        for(int i=1;i<=nums.size();i++)
        {
            x2^=i;
        }
        result = x1^x2;
        return result;
    }
};