268. Missing Number -- 找出0-n中缺失的一个数

 

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?

（1）

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

异或0-n，异或nums。

异或0还等于原来的数。

（2）

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int n = nums.size(), sum = n*(n+1) / 2;
        for(int i = 0; i < n; i++)
        {
            sum -= nums[i];
        }
        return sum;
    }
};

求和相减。