1. 原题链接 https://leetcode.com/problems/swap-nodes-in-pairs/description/ 2. 题目要求 给定一个链表,交换相邻的两个结点.已经交换的结点,不再进行交换. 注意:所使用的空间大小固定 例如,1->2->3->4转换后为2->1->4->3 3. 题目思路 使用一个遍历指针current和两个辅助指针first.second,first保存current指针所在结点的后继结点,second保存curren…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 24: Swap Nodes in Pairshttps://oj.leetcode.com/problems/swap-nodes-in-pairs/ Given a linked list, swap every two adjacent nodes and return its head. For example,Given 1->2->3->4, you sh…

Swap Nodes in Pairs Given a linked list, swap every two adjacent nodes and return its head. For example,Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the val…

一天一道LeetCode系列 (一)题目 Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the v…

题目链接: https://leetcode.com/problems/swap-nodes-in-pairs/?tab=Description   Problem: 交换相邻的两个节点     如上图所示,递归进行交换.从最尾端开始,当最尾端只有一个节点时,停止交换 否则执行 swap(head.next)    参考代码:   package leetcode_50; /** * * @author pengfei_zheng * 交换相邻节点 */ public class Solutio…

题意:交换链表中每两个相邻节点,不能修改节点的val值. 分析:递归.如果以第三个结点为头结点的链表已经两两交换完毕(这一步递归实现---swapPairs(head -> next -> next)),则接下来,只需交换前两个节点,再将第一个节点的next指向“以第三个结点为头结点的链表两两交换后”的链表. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * L…

Given a linked list, swap every two adjacent nodes and return its head. For example,Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the values in the list, onl…

Given a linked list, swap every two adjacent nodes and return its head. For example,Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the values in the list, onl…

Given a linked list, swap every two adjacent nodes and return its head. For example,Given 1->2->3->4, you should return the list as 2->1->4->3. Your algorithm should use only constant space. You may not modify the values in the list, onl…

不允许通过值来交换,在更新指针时需要小心. ListNode *swapNodes(ListNode* head) { ListNode dummy(-); dummy.next = head; for (ListNode *prev = &dummy, *cur = prev->next, *pnext = cur->next; pnext; prev = cur, cur = cur->next, pnext = cur->next) { prev->next =…