260.Single Number III

最新推荐文章于 2024-07-25 07:21:48 发布

原创最新推荐文章于 2024-07-25 07:21:48 发布 · 420 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #算法

leetcode OJ 专栏收录该内容

15 篇文章

订阅专栏

本文介绍了一种算法，用于在一个包含重复元素的数组中找到只出现一次的两个元素。通过使用HashSet的数据结构，该算法在O(n)时间内运行且仅使用常数空间。实例演示了如何应用此算法。

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?

public class Solution {
    /*
     * 利用HashSet的唯一性
     */
    public int[] singleNumber(int[] nums) {
        HashSet<Integer> hash = new HashSet<>();
        for(int n: nums){
            if(!hash.add(n)){
                hash.remove(n);
            }
        }
        int i = 0;
        int[] array = new int[hash.size()];
        for(int b: hash){
            array[i++] = b;
        }
        return array;
    }
}