Array-Find All Duplicates in an Array

最新推荐文章于 2025-06-06 12:39:01 发布

kevin聪

最新推荐文章于 2025-06-06 12:39:01 发布

阅读量183

点赞数

CC 4.0 BY-SA版权

分类专栏：算法与数据结构 leetcode 文章标签：算法数据结构 leetcode

本文链接：https://blog.youkuaiyun.com/LaputaFallen/article/details/78649998

算法与数据结构同时被 2 个专栏收录

569 篇文章

订阅专栏

leetcode

557 篇文章

订阅专栏

本文介绍了一种高效查找数组中重复元素的方法，不使用额外空间且运行时间为O(n)。通过两个解决方案对比，展示了如何利用数组特性快速定位重复项。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description:

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]

My Solution:

//不用Arrays.sort(),用数组存储元素对应的个数会更快
class Solution {
    public List<Integer> findDuplicates(int[] nums) {
        java.util.Arrays.sort(nums);
        List<Integer> list = new ArrayList<Integer>();
        int len = nums.length;
        for(int i = 0;i < len - 1;i++){
            if(nums[i + 1] == nums[i]){
                list.add(nums[i]);
                i++;
            }
        }
        return list;
    }
}

Better Solution:

public class Solution {
    // when find a number i, flip the number at position i-1 to negative. 
    // if the number at position i-1 is already negative, i is the number that occurs twice.

    public List<Integer> findDuplicates(int[] nums) {
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < nums.length; ++i) {
            int index = Math.abs(nums[i])-1;
            if (nums[index] < 0)
                res.add(Math.abs(index+1));
            nums[index] = -nums[index];
        }
        return res;
    }
}