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 may not modify the values in the list, only nodes itself can be...
代码如下: 1/**2* Definition for singly-linked list.3* public class ListNode {4* int val;5* ListNode next;6* ListNode(int x) { val = x; }7* }8*/9publicclassSolution {10publicListNode swapPairs(ListNode head) {11if(head ==null)returnhead;12//一般需要处理head的时候会用到dumpnode,...
leetcode 24. 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 values in the list, only n...
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. not answer: class Solution { public: ListNode* swapPairs(ListNode* head) { ListNode* first = head; ListNode* second ; if(first == NU...
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] ...
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
这道题目没什么好说的,虽说是medium,但是挺简单。感觉写过insertion sort list 之后,链表类的简单题,中等题都不虚了。。。 感觉代码写的还不是很好。让人感觉太多判断,不顺其自然。 改写了下,如下: publicListNodeswapPairs(ListNodehead){if(head==null||head.next==null)returnhead;ListNodedummy=newListNode(...
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...
* ListNode(int x) { val = x; } * } */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...