203. 移除链表元素
给你一个链表的头节点head
和一个整数 val
,请你删除链表中所有满足 Node.val == val
的节点,并返回 新的头节点 。
示例 1:
![](https://assets.leetcode.com/uploads/2021/03/06/removelinked-list.jpg)
输入:head = [1,2,6,3,4,5,6], val = 6 输出:[1,2,3,4,5]
示例 2:
输入:head = [], val = 1 输出:[]
示例 3:
输入:head = [7,7,7,7], val = 7 输出:[]
提示:
- 列表中的节点数目在范围
[0, 104]
内 1 <= Node.val <= 50
0 <= val <= 50
解法一:虚拟头结点(易于理解)
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode removeElements(ListNode head, int val) { if (head == null) { return head; } ListNode dummy = new ListNode(-1, head); ListNode pre = dummy; ListNode cur = head; while (cur != null) { if (cur.val == val) { pre.next = cur.next; } else { pre = cur; } cur = cur.next; } return dummy.next; } }
解法二:递归
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ class Solution { public ListNode removeElements(ListNode head, int val) { if (head == null) { return head; } head.next = removeElements(head.next, val); return head.val == val ? head.next : head; } }
标签:203,ListNode,cur,val,int,head,next,链表,移除 From: https://www.cnblogs.com/fulaien/p/16870491.html