[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?

首先对0～ n内数字做异或，在对数组内数字做异或。等价与在Single Number 1，在一堆重复数字中找1个。

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