1.5 KiB
1.5 KiB
删除链表的倒数第 N 个结点
题目
给你一个链表,删除链表的倒数第 n 个结点,并且返回链表的头结点。
示例 1:
输入:head = [1,2,3,4,5], n = 2 输出:[1,2,3,5] 示例 2:
输入:head = [1], n = 1 输出:[] 示例 3:
输入:head = [1,2], n = 1 输出:[1]
提示:
链表中结点的数目为 sz 1 <= sz <= 30 0 <= Node.val <= 100 1 <= n <= sz
进阶:你能尝试使用一趟扫描实现吗?
思路
- 双指针
- 快指针指向 null
- 慢指针指向目标结点
代码
/**
* 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 removeNthFromEnd(ListNode head, int n) {
// Init: slow, fast, dummyHead
ListNode dummyHead = new ListNode(-1);
dummyHead.next = head;
ListNode slow = dummyHead;
ListNode fast = dummyHead;
int cnt = 0;
while (fast != null && cnt < n) {
fast = fast.next;
cnt++;
}
if (cnt != n) {
return null;
}
while (fast != null) {
fast = fast.next;
if (fast == null) {
slow.next = slow.next.next;
return dummyHead.next;
}
slow = slow.next;
}
return dummyHead.next;
}
}