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 。


- /**
- * 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;
- }
- }