Given1->2->3->4, you should return the list as2->1->4->3. Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed. 代码:oj测试164ms通过 1#Definition for singly-linked list.2#class ListNode:3#def __init__(self,...
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] ...
LeetCode(24) - Swap Nodes in Pairs 题目要求很简单,就是把list里面的node两两互换。 当做比较复杂的LinkedList的题目的时候,最好是在草稿纸上画一画,走一遍流程,凭空想很容易会出错。这一题时LeetCode 25的k=2特殊例子,所以要简单很多。用两个node,一前一后,然后两两交换就好,细节上注意的是交换后两个nod...
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] Output: [2,1,4,3] Explanation: Example 2: Inpu...
【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. ...
The number of nodes in the list is in the range[0, 100]. 0 <= Node.val <= 100 两两交换链表中的节点。给一个linked list,请成对调换node。 我的思路是迭代。依然是给一个dummy节点放在head节点之前,然后dummy.next是head节点,nextStart是第三个节点,需要交换的只是l2和l2.next,l1永远是需要swap的...
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
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. ...
Leetcode-Swap Nodes in Pairs 一、题目——Swap Nodes in Pairs Given a linked list, swap every two adjacent nodes and return its head. For example, Given1->2->3->4, you should return the list as2->1->4->3. Your algorithm should use only constant space. You maynotmodify the values...
*//** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */publicclassSolution{/** * @param head a ListNode * @return a ListNode */publicListNodeswapPairs(ListNodehead){ListNodedummy=newListNode(0);dummy...