【LeetCode OJ 260】Single Number III

题目链接：https://leetcode.com/problems/single-number-iii/

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?

解题思路：题意为给定一个数组，只有两个数字出现了一次，其它均出现了两次，请找出这两个出现了一次的数。题目要求算法的时间复杂度为线性的，空间复杂度为常量，本题给出了一个粗暴的解法：
代码示例：

public class Solution 
{
	 public int[] singleNumber(int[] nums)
	 {
		 int[] result=new int[2];
		 List<Integer> temp=new ArrayList<Integer>();
		 for(int i=0;i<nums.length;i++)
		 {
			 //如果不存在，则加入temp中
			 if(!temp.contains(nums[i]))
			 {
				 temp.add(nums[i]);
			 }
			 //不存在，表示存在两次，就从temp除去该数
			 else
			 {
				 temp.remove((Object)nums[i]);
				 
			 }
		 }
		 result[0]=temp.get(0);
		 result[1]=temp.get(1);
		 return result;
	 }
}