Leet Code OJ 260. Single Number III [Difficulty: Medium]

题目：
Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:
The order of the result is not important. So in the above example, [5, 3] is also correct.
Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

翻译：
给定一个数组，里面有2个元素，只出现一次，其他元素都是出现2次。求这2个出现一次的元素。
说明：返回结果的数组顺序不重要。你的程序需要运行在线性复杂度下。你可以实现它只用常数级的空间复杂度吗？

代码：

public class Solution {public int[] singleNumber(int[] nums) {int[] result=new int[2];Map<Integer,Integer> map=new HashMap<>();for(int i=0;i<nums.length;i++){Integer value=map.get(nums[i]);if(value==null){map.put(nums[i],1);}else{map.put(nums[i],1+value);}}int resultIndex=0;for(Integer key:map.keySet()){if(map.get(key)==1){result[resultIndex]=key;resultIndex++;}}return result;}
}

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