[LeetCode]Single Number

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?

1. 不考虑多余内存空间，用HashMap解决很常规，时间复杂度为O(n)。

public class Solution {    public int singleNumber(int[] A) {        HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();        int key,val;        val=0;        for(int i=0;i<A.length;i++){            if(map.containsKey(A[i])){                map.remove(A[i]);            }            else{                map.put(A[i],A[i]);            }        }        Iterator iter=map.keySet().iterator();        while(iter.hasNext()){            val=map.get(iter.next());        }        return val;    }}

2. note里又说do it without using extra memory，这就难办了。想了很久没找到诀窍，最后求助万能互联网。

异或运算：^

输入输入结果110101011000

代码变得十分简洁，太棒了！

public class Solution {    public int singleNumber(int[] A) {        for(int i=1;i<A.length;i++){           A[0]=A[0]^A[i];        }        return A[0];    }}