Leetcode 92题反转链表 II(Reverse Linked List II) 反转链表可以先看这篇文章:LeetCode 206题 反转链表(Reverse Linked List) 题目链接 https://leetcode-cn.com/problems/reverse-linked-list-ii/ 题目描述 反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。 说明: 1 ≤ m ≤ n ≤ 链表长度。 示例:...
此解法与第四种解法思路类似,只不过是将栈换成了数组,然后新建node节点,以数组最后一位元素作为节点值,然后开始循环处理每个新的节点。 publicListNodereverseList5(ListNode head){if(head ==null|| head.next ==null) {returnhead; } ArrayList<Integer> list =newArrayList<Integer>();while(head !=null) { ...
解法二(Java) /*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * }*/classSolution {publicListNode reverseList(ListNode head) {if(head ==null|| head.next ==null)returnhead;//处理最小输入的情况,即空链表...
英文网站:92. Reverse Linked List II 中文网站:92. 反转链表 II 问题描述 Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list. Example 1: Input: head ...
leetcode 206. Reverse Linked List 反转字符串,Reverseasinglylinkedlist.反转链表,我这里是采用头插法来实现反转链表。代码如下:/*classListNode{intval;ListNodenext;ListNode(intx){val=x;}}*/publicclassSolution{publicListNoderever
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: ...
LeetCode 206. 反转链表(Reverse Linked List) 示例: 输入:1->2->3->4->5->NULL输出:5->4->3->2->1->NULL 切题 一、Clarification 只需注意为空链表的情况 二、Possible Solution 1、迭代 2、递归 可利用哨兵简化实现难度 Python3 实现
LeetCode-链表 链表(Linked List)是一种常见的基础数据结构,是一种线性表,但是并不会按线性的顺... raincoffee阅读 1,218评论 0赞 6 Linked List的复习总结 Single Linked List 相比较另一个基本的数据结构array,linked list有几个优势:尺寸... dol_re_mi阅读 8,197评论 0赞 3 穿越到金庸武侠世界中的囧...
Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 递归法 复杂度 时间O(N) 空间 O(N) 递归栈空间 思路 基本递归 代码 public class Solution { ...
Leetcode260反转链表(java/c++/python) JAVA: class Solution { public ListNode reverseList(ListNode head) { if( head == null || head.next == null) return head; ListNode newHead = reverseList(head.next); head.next.next = head; head.next = null; return newHead; } } C++: class Solution...