leetcode single-number

Posted 2020-10-04

题目描述

 

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?

 

我的解题思路：不能使用额外空间，时间要线性，粗暴的方法就是两个循环

 

在下的代码 时间213ms 空间13164k

public class Solution {
    public int singleNumber(int[] A) {
        for (int i=0;i<A.length;i++){
            Boolean twice = false;
            for (int j= 0;j<A.length;j++){
                if (i == j)
                    continue;
                if (A[i] == A[j]){
                    twice = true;
                    break;
                }
            }
            if (!twice){
                return A[i];
            }
        }
        return 0;
    }
}

　　

大神代码 时间150ms 空间12952k 异或的思路就是好啊，两个相同的数异或为0

public class Solution {
    public int singleNumber(int[] A) {
        int num = 0;
        for(int i=0;i<A.length;i++){
            num^=A[i];
        }
        return num;
    }
}