【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:

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

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;
	 }
}

【LeetCode OJ 260】Single Number III

原文：http://blog.csdn.net/xujian_2014/article/details/50581165