203.移除链表元素
题目描述
原题
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:
- 列表中的节点在范围
[0, 104]
内
1 <= Node.val <= 50
0 <= k <= 50
题解
递归
| /**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
if(head == null)
return null;
while(head!=null && head.val==val){
head = head.next;
}
if(head!=null)
head.next = removeElements(head.next,val);
return head;
}
}
|
迭代
| /**
* 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) {
//创建一个模拟的头节点
ListNode dummyNode = new ListNode();
dummyNode.next = head;
head = dummyNode;
while(head!=null){
ListNode next = head.next;
//循环找到值不相等的next节点
while(next!=null&&next.val == val){
next = next.next;
}
//指向next
head.next = next;
head = next;
}
return dummyNode.next;
}
}
|
| //官方代码优化
class Solution {
public ListNode removeElements(ListNode head, int val) {
ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode temp = dummyHead;
while (temp.next != null) {
if (temp.next.val == val) {
temp.next = temp.next.next;
} else {
temp = temp.next;
}
}
return dummyHead.next;
}
}
|