Leetcode86: 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.

根据异或的特性，对于一个数，异或自己是0，异或0是自己，所以我们把0-n对着给定数组异或一遍，结果就是缺失的数。

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