《leetcode》single-number

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?

解析：该题目暴力求解当然是可以通过的，但是不满足题目的要求，题目要求时间复杂度为O（n）。

代码一：暴力求解，不建议

import java.util.*;
public class Solution {public int singleNumber(int[] A) {List<Integer> list = new ArrayList<>();for(int i:A){list.add(i);}Set<Integer> set = new HashSet<>(list);for(int i:set){int temp =i;int count=0;for(int j:A){if(j==i){count++;}}if(count==1){return temp;}}return -1;}
}

代码二：利用异或的计算特点。
1）相同的数异或为0,例如 6^6=0；
2）0异或不为0的数为不为0的数，例如：0^5=5

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

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