Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1

Input: [3,0,1]
Output: 2

 

Example 2

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

 

 

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.

常规方法，直接上代码：

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        
        int n = nums.size();
        
        return (0 + n)*(n + 1) / 2 - accumulate(nums.begin(), nums.end(), 0);
    }
};

还有一种就是用异或XOR的性质了，X^X=0, X^0=X，满足交换律：

class Solution {
public:
    int missingNumber(vector<int>& nums) {

        int n = nums.size();
        int res = n;
        for (int i = 0; i < n; ++i)
        {
            res ^= nums[i] ^ i;
        }
        return res;
    }
};