Swap Nodes in Pairs 思路:需要构造一个头指针,指向链表。一次比较两个指针,将其翻转,临界条件是pre != null(链表长度为偶数) && pre.next != null(链表长度为奇数),然后更新头指针和pre publicListNodeswapPairs(ListNode head){if(head ==null)returnnull;ListNodenewNode=newListNode(0);ListNodenode=newNode...
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...
先寻找出正数第k个节点和倒数第k个节点,然后交换二者的值,间接实现了交换节点的目的。 代码: class Solution { public ListNode swapNodes(ListNode head, int k) { ListNode A = head; //正数第k个节点A ListNode B = head; //倒数第k个节点B ListNode nodeK; //用来保存正数第k个节点A // step1: ...
依据这个信息,我们用两个指针从head节点开始遍历,分别在两个被swap的节点停下,swap他们的节点值即可。 时间O(n) 空间O(1) Java实现 1/**2* Definition for singly-linked list.3* public class ListNode {4* int val;5* ListNode next;6* ListNode() {}7* ListNode(int val) { this.val = val; }...
817. Linked List Components题解 也可以用递归的方式遍历链表,但这种方式会重复访问节点,时间复杂度比O(n)高很多。 相关LeetCode题: 24. Swap Nodes in Pairs题解 234. Palindrome Linked List题解 反转/旋转链表 反转(reverse)链表与旋转(rotate)链表是考量链表操作的经典题目,最是考验做题人对链表节点逻辑关系...
The number of nodes in the list isn. 1 <= k <= n <= 105 0 <= Node.val <= 100 classSolution {publicListNode swapNodes(ListNode head,intk) { ListNode dummy=newListNode(-1); dummy.next=head; ListNode p= dummy, q =dummy;for(inti =0; i < k; i++) p =p.next;while(p.next...
思路:最简单的就是建一个Dummy node,然后不断地将原来List的Node插入到dummy node的后面, 但是这样需要了额外的空间。 更好的方法是用一个variable pre保存前一个node,一个cur保存现在的Node,不断地改变这两个node 的指针关系,并且将pre和cur更新向下两个点。
1.) The nodes may not be adjacent to each other 2.) The nodes may be the first or last nodes in the list 3.) The nodes may be out of order (I assume the first one comes before the second one, but the user may pass them in reverse order) ...
24 Swap Nodes in Pairs classSolution:defswapPairs(self,head:ListNode)->ListNode:ifhead isNoneorhead.nextisNone:returnhead realhead=ListNode(0)realhead.next=head cur=realheadwhile(cur.nextandcur.next.next):first=cur.nextsecond=cur.next.nextfirst.next=second.nextcur.next=second second.next=firs...
Swap Nodes in Pairs Add Two Numbers You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. ...