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This article mainly introduces how to solve the problem of finding the last-to-last node of the leetcode linked list, which has a certain reference value, and interested friends can refer to it. I hope you can learn a lot after reading this article.

Enter a linked list and output the last-to-last node in the linked list. In order to meet the habits of most people, this topic starts counting from 1, that is, the end node of the linked list is the penultimate node. For example, a linked list has six nodes, starting from the top node, and their values are 1, 2, 3, 4, 5, and 6. The penultimate node of this linked list is the node with a value of 4. Example: given a linked list: 1-> 2-> 3-> 4-> 5, and k = 2. Return to linked list 4-> 5. Source: LeetCode link: https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof copyright belongs to the collar buckle network. For commercial reprint, please contact official authorization. For non-commercial reprint, please indicate the source. Answer the question / Definition for singly-linked list. * public class ListNode {* int val; * ListNode next; * ListNode (int x) {val = x;} *} * / class Solution {public ListNode getKthFromEnd (ListNode head, int k) {ListNode C1 = head; while (C1! = null & & k > 0) {C1 = c1.nextKmurf;} ListNode c2 = head While (C1! = null) {C1 = c1.next; c2 = c2.next;} return c2;}}

Fast and slow pointer, first let the fast pointer take k steps, and then the two pointers go synchronously. When the fast pointer reaches the end, the slow pointer is the last-to-last node of the linked list.

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