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 · 614 阅读

0 ·

CC 4.0 BY-SA版权

Linked List 专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一种从已排序链表中移除所有具有重复数值的节点的方法，只保留原始列表中的不同数值。通过示例展示了如何将1->2->3->3->4->4->5转换为1->2->5，以及将1->1->1->2->3转换为2->3。

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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if (head == null)
			return head;
		ListNode res = new ListNode(0);
		res.next = head;
		ListNode pre = res;
		ListNode node = head;
		ListNode next = node.next;
		while (next != null) {
			if (node.val == next.val) {
				while (next != null && node.val == next.val) {
					next = next.next;
				}
				pre.next = next;
				node = next;
				if(next != null)
					next = next.next;
			} else {
				pre = pre.next;
				node = node.next;
				next = next.next;
			}
		}
		return res.next;
	}
}