LeetCode- 19 删除链表的倒数第N个节点

题目地址

https://leetcode.cn/problems/remove-nth-node-from-end-of-list/description/

参考实现

 /*** 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; }* }*//*** 通过链表长度 进行处理** @param head* @param n* @return*/public static ListNode removeNthFromEnd1(ListNode head, int n) {ListNode dumny = new ListNode(0, head);ListNode current = dumny;for (int i = 1; i < getLenth1(head) - n + 1; i++) {current = current.next;}//执行删除current.next = current.next.next;return dumny.next;}/*** 获取链表大小** @param head* @return*/public static int getLenth1(ListNode head) {int length = 0;while (head != null) {head = head.next;length++;}return length;}/*** 利用栈特性处理** @param head* @param n* @return*/public static ListNode removeNthFromEnd(ListNode head, int n) {ListNode dumny = new ListNode(0, head);Deque<ListNode> stack = (Deque<ListNode>) new LinkedList<ListNode>();ListNode current = dumny;while (current != null) {stack.push(current);current = current.next;}for (int i = 0; i < n; i++) {stack.pop();}ListNode peek = stack.peek();peek.next = peek.next.next;return dumny.next;}/*** 双指针处理** @param head* @param n* @return*/public static ListNode removeNthFromEnd3(ListNode head, int n) {ListNode dummy = new ListNode(0, head);ListNode first = head;ListNode second = dummy;for (int i = 0; i < n; i++) {first = first.next;}while (first != null) {first = first.next;second = second.next;}second.next = second.next.next;return dummy.next;}