Remove Duplicates from Sorted List II

最新推荐文章于 2024-01-17 14:07:10 发布

原创最新推荐文章于 2024-01-17 14:07:10 发布 · 215 阅读

0 ·

CC 4.0 BY-SA版权

算法专栏收录该内容

131 篇文章

订阅专栏

本文介绍如何在保持链表排序的情况下，移除所有重复的节点，仅保留唯一的数值。通过实例演示了如何实现这一功能，包括核心算法步骤及应用示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Have you met this question in a real interview? Yes
Example
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.


public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null)
            return head;

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;

        while (head.next != null && head.next.next != null) {
            if (head.next.val == head.next.next.val) {
                int val = head.next.val;
                while (head.next != null && head.next.val == val) {
                    head.next = head.next.next;
                }            
            } else {
                head = head.next;
            }
        }

        return dummy.next;
    }
}