Find All Duplicates in An Array

最新推荐文章于 2025-09-08 22:54:36 发布

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最新推荐文章于 2025-09-08 22:54:36 发布

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CC 4.0 BY-SA版权

文章标签： runtime

原文链接：http://www.cnblogs.com/amazingzoe/p/6411699.html

本文介绍了一种在整数数组中查找所有出现两次的元素的方法。该算法能够在O(n)的时间复杂度内完成，并且不使用额外的空间。通过遍历数组并进行交换来检测重复项，最终返回所有找到的重复元素。

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements that appear twice in this array.

Could you do it without extra space and in O(n) runtime?

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[2,3]

 1 public class Solution {
 2     public List<Integer> findDuplicates(int[] nums) {
 3         List<Integer> result = new ArrayList<Integer>();
 4         
 5         for (int i = 0; i < nums.length; i++) {
 6             while(i + 1 != nums[i] && nums[i] != -1) {
 7                 if (nums[i] == nums[nums[i] - 1]) {
 8                     result.add(nums[i]);
 9                     nums[i] = nums[nums[i] - 1] = -1;
10                     break;
11                 }   
12                 else swap(nums, i, nums[i] - 1);
13             }
14         }
15         return result;
16     }
17     
18     private void swap(int[]nums, int i, int j) {
19         int temp = nums[i];
20         nums[i] = nums[j];
21         nums[j] = temp;
22     }
23 }