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:
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:
- 列表中的节点在范围
[0, 104]
内 1 <= Node.val <= 50
0 <= k <= 50
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeElements(self, head: ListNode, val: int) -> ListNode:
dummy = ListNode(-1, head)
pre = dummy
while pre and pre.next:
if pre.next.val != val:
pre = pre.next
else:
pre.next = pre.next.next
return dummy.next
/**
* 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) {
ListNode dummy = new ListNode(-1, head);
ListNode pre = dummy;
while (pre != null && pre.next != null) {
if (pre.next.val != val) pre = pre.next;
else pre.next = pre.next.next;
}
return dummy.next;
}
}