LeeCode_136 Single Number

Given an array of integers, every element appears twice except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

考虑使用异或运算来实现，异或运算为两个数各bit为一一比较，如果相同置为0，如果不同置为1.

两个性质：

a^b=b^a;

0^a=a;

所以只需将数列不断的对res做异或运算，肯定可以出现（n-1）/2个0， 还剩最后一个就是要找的数。

class Solution {
public:
int singleNumber(vector<int>& nums) {

int res=0;

for (int i=0;i<nums.size();i++)
res=res^nums[i];
return res;

}
};