LeetCode *** 268. Missing Number

题目：

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?

分析：

嗯，这道题是从一堆数字里找到失去的那个，那么可以用到位操作中的异或，先得到0~n所有数字的异或，然后将该结果分别与nums[i]进行异或。这样最后剩下的数字就是丢失的那个。

代码：

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