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?
题解:
方法:双指针法
利用两个指针,给两个指针设置N + 1的间距,那么当快的哪个指针指到最后一个节点的下一个节点(即null)时,慢指针会指到删除目标的前一个位置
图解
代码演示:
/*** 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) {ListNode dumyNode = new ListNode(0);dumyNode.next = head;ListNode fastIndex = dumyNode;ListNode slowIndex = dumyNode;for(int i = 0; i < n + 1;i++) {fastIndex = fastIndex.next;}while(fastIndex != null){slowIndex = slowIndex.next;fastIndex = fastIndex.next;}slowIndex.next = slowIndex.next.next;return dumyNode.next;}
}