LeetCode:Single Number

标签：

Problems:

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:这里主要利用的是异或运算的性质，出现偶数次则x清零。基数次保留结果。

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        //考察的是位运算        
        int x=0;
        
        for(int i=0;i<nums.size();i++)
            x^=nums[i];
        return x;
    }
};

LeetCode:Single Number

标签：

原文地址：http://www.cnblogs.com/xiaoying1245970347/p/4639254.html