LeetCode 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.

非常妙的一题
使用xor操作, nums中给出的n个数其范围应该在[0, n]之间（共n+1个）即0.1.2.3...n，然后抽掉了一个数，形成nums中的数字（共n个）。所以使用0.1.2.3...n这n+1个数字和数组内各个元素做xor，n+n+1= 2n+1，这样下来必有一个数找不到和他一致的xor为零的数，这个数就是nums数组中缺少的那个

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

或者也可以使用数列求和公式

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int len = nums.size();
        
        double sum_should = (0 + len) * (len + 1) / 2.0;
        double sum_actual = 0;
        for (int e : nums) {
            sum_actual += e;
        }
        return sum_should - sum_actual;
    }
};