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?

题意：给出数组nums，nums的长度是n，里边放的是0-n之间的不同的值，找出丢失的值。

思路：把数组所有值加一块儿sum，用0-n的总和值减去sum就是缺失的值。

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

思路2：大神的思路是，假设0-5之间的缺失值是4，则0^1^2^3^4^5^0^1^2^3^4^5=4.利用了0^n=n, n^n=0的性质。

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