## 题目地址 https://leetcode.com/problems/remove-nth-node-from-end-of-list/description ## 题目描述 Given a linked list, remove the n-th node from the end of list and return its head. Example: Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5. Note: Given n will always be valid. Follow up: Could you do this in one pass? ## 思路 双指针,指针A先移动n次, 指针B再开始移动。当A到达null的时候, 指针b的位置正好是倒数n 我们可以设想假设设定了双指针p和q的话,当q指向末尾的NULL,p与q之间相隔的元素个数为n时,那么删除掉p的下一个指针就完成了要求。 设置虚拟节点dummyHead指向head 设定双指针p和q,初始都指向虚拟节点dummyHead 移动q,直到p与q之间相隔的元素个数为n 同时移动p与q,直到q指向的为NULL 将p的下一个节点指向下下个节点 ![19.removeNthNodeFromEndOfList](../assets/19.removeNthNodeFromEndOfList.gif) (图片来自: https://github.com/MisterBooo/LeetCodeAnimation) ## 关键点解析 1. 链表这种数据结构的特点和使用 2. 使用双指针 3. 使用一个dummyHead简化操作 ## 代码 Support: JS, Java - Javascript Implementation ```js /* * @lc app=leetcode id=19 lang=javascript * * [19] Remove Nth Node From End of List * * https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/ * * algorithms * Medium (34.03%) * Total Accepted: 360.1K * Total Submissions: 1.1M * Testcase Example: '[1,2,3,4,5]\n2' * * Given a linked list, remove the n-th node from the end of list and return * its head. * * Example: * * * Given linked list: 1->2->3->4->5, and n = 2. * * After removing the second node from the end, the linked list becomes * 1->2->3->5. * * * Note: * * Given n will always be valid. * * Follow up: * * Could you do this in one pass? * */ /** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} head * @param {number} n * @return {ListNode} */ var removeNthFromEnd = function(head, n) { let i = -1; const noop = { next: null }; const dummyHead = new ListNode(); // 增加一个dummyHead 简化操作 dummyHead.next = head; let currentP1 = dummyHead; let currentP2 = dummyHead; while (currentP1) { if (i === n) { currentP2 = currentP2.next; } if (i !== n) { i++; } currentP1 = currentP1.next; } currentP2.next = ((currentP2 || noop).next || noop).next; return dummyHead.next; }; ``` - Java Code ```java /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { TreeNode dummy = new TreeNode(0); dummy.next = head; TreeNode first = dummy; TreeNode second = dummy; if (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; } } ```