19. 删除链表的倒数第 N 个结点
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?
/**
* 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 {
/**
快慢指针应用:快指针移到尾元素,慢指针下一个节点为目标节点,便于删除。
加入logos前指针便于处理单元素删除问题。
*/
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode logos = new ListNode(0, head);
ListNode slow = logos, fast = logos;
for(int i = 0; i < n; i++){
fast = fast.next;
}
while(fast.next != null){
slow = slow.next;
fast = fast.next;
}
ListNode target = slow.next;
slow.next = target.next;
target.next = null;
return logos.next;
}
}