Remove Duplicates from Sorted List II 解答

最新推荐文章于 2024-01-17 14:07:10 发布

转载最新推荐文章于 2024-01-17 14:07:10 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/ireneyanglan/p/4814867.html

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，包括特殊情况下如整个链表元素相同的情况。

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Question

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

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

Solution

Note here we need to consider head of the list. For example, for input 1 -> 1, we need to return null.

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode deleteDuplicates(ListNode head) {
11         if (head == null || head.next == null)
12             return head;
13         while (head.next != null && head.val == head.next.val) {
14             while (head.next != null && head.val == head.next.val)
15                 head = head.next;
16             if (head.next != null)
17                 head = head.next;
18             else
19                 return null;
20         }
21         ListNode current = head, next1, next2;
22         while (current != null && current.next != null && current.next.next != null) {
23             next1 = current.next;
24             next2 = current.next.next;
25             if (next1.val != next2.val) {
26                 current = current.next;
27             } else {
28                 while (next2 != null && next1.val == next2.val)
29                     next2 = next2.next;
30                 current.next = next2;
31             }
32         }
33         return head;
34     }
35 }

转载于:https://www.cnblogs.com/ireneyanglan/p/4814867.html