[Leetcode]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?

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

还可以用位运算来做这个题，利用a ^ a = 0  a ^ b ^ a = b;

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