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;
}
};