leetcode-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?

思路：遍历异或，最后剩下的就是唯一的那个数

class Solution {public:    int singleNumber(vector<int>& nums) {        //依次遍历异或，最后剩的就是唯一的那个值        //扩展：如果有两个不同的，异或完后看结果的哪一个bit为1，        //选一个，根据这个bit为1和0把数组分成两组，分别遍历异或得到两个不同值        int length = nums.size();        if(length == 0)        {            return 0;        }        int result = nums[0];        for(int i = 1;i < length;i++)        {            result = result^nums[i];        }        return result;    }};