classSolution{publicListNodeswapPairs(ListNode head){ListNodedummy=newListNode(0); dummy.next = head; ListNode pre= dummy;while(pre.next !=null&& pre.next.next !=null){ swap(pre); pre = pre.next.next; }returndummy.next; }publicvoidswap(ListNode pre){ListNodenext=pre.next; pre.next = n...
* }*/classSolution {publicListNode swapPairs(ListNode head) {if(head ==null|| head.next ==null)returnhead; ListNode second= head.next;//存储第二个节点在second中head.next = swapPairs(second.next);//要递归处理second节点后的那个节点,并将翻转好的结果链接到second的前驱后,即headsecond.next = ...
【LeetCode】Swap Nodes in Pairs 解题报告 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/swap-nodes-in-pairs/description/ 题目描述: Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2->3->4, you should return ...
给你一个链表,两两交换其中相邻的节点,并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题(即,只能进行节点交换)。 示例1: 输入:head = [1,2,3,4]输出:[2,1,4,3] 示例2: 输入:head = []输出:[] 示例3: 输入:head = [1]输出:[1] ...
24. Swap Nodes in Pairs Medium Topics Companies Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.) Example 1: Input: head = [1,2,3,4] ...
Leetcode 24.Swap Nodes in Pairs 给你一个链表,交换相邻两个节点,例如给你 1->2->3->4,输出2->1->4->3。 我代码里在head之前新增了一个节点newhead,其实是为了少写一些判断head的代码。 public class Solution { public ListNode swapPairs(ListNode head) {...
Can you solve this real interview question? Swap Nodes in Pairs - Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be chan
1、迭代实现 Java 两两交换其中相邻的节点(Swap Nodes in Pairs) Java 迭代实现 2、递归实现 Java 两两交换其中相邻的节点(Swap Nodes in Pairs) Java 递归实现 GitHub链接: https://github.com/lichangke/LeetCode 知乎个人首页: https://www.zhihu.com/people/lichangke/ ...
总结: Linkedlist, swap nodes ** Anyway, Good luck, Richardo! 其实就是 Reverse Nodes in k-Group - http://www.jianshu.com/p/25f5269c729d 的特殊情况。 思路很简单,没什么好说的。 My code: /** * Definition for singly-linked list. ...
*/publicclassSolution{/** * @param head a ListNode * @return a ListNode */publicListNodeswapPairs(ListNodehead){ListNodedummy=newListNode(0);dummy.next=head;head=dummy;while(head.next!=null&&head.next.next!=null){ListNoden1=head.next,n2=head.next.next;// head->n1->n2->...// => he...