https://leetcode-cn.com/problems/reverse-nodes-in-k-group/description/ 给出一个链表,每 k 个节点一组进行翻转,并返回翻转后的链表。 k 是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍,那么将最后剩余节点保持原有顺序。 解法1 首先遍历链表,记录结点个数,当满足个数等于k时,...
1. Reverse Nodes in k-Group Given a linked list, reverse the nodes of a linked listkat a time and return its modified list. kis a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple ofkthen left-out nodes in the...
Given a linked list, reverse the nodes of a linked listkat a time and return its modified list. kis a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple ofkthen left-out nodes in the end should remain as it is....
[LeetCode] 025. Reverse Nodes in k-Group (Hard) (C++/Java),有需要的朋友可以参考下。 索引:[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql) Github:https:///illuz/leetcode 025. Reverse Nodes in k-Group (Hard) 链接: 题目:https://oj.leetcode.com/problems/reverse-nodes-in-k-group/...
LeetCode.jpg 目录链接:https://www.jianshu.com/p/9c0ada9e0ede k个一组翻转链表 给出一个链表,每 k 个节点一组进行翻转,并返回翻转后的链表。 k 是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍,那么将最后剩余节点保持原有顺序。
:type k: int :rtype: ListNode """p=head stack=[]m=kwhilepandm:stack.append(p)m-=1p=p.nextifm>0:returnhead newhead=ListNode(0)q=newheadwhilestack:q.next=stack.pop()q=q.nextq.next=self.reverseKGroup(p,k)returnnewhead.next ...
Reverse Nodes in k-Group@LeetCode Reverse Nodes in k-Group 翻转链表的升级版,由于在以前阿里的面试中遇到过,所以特别在意,这次写题解之前又重新把原来通过的代码优化了一下。 其实这题很简单,知道了翻转链表的通用写法之后,解这一题其实就是循环翻转的过程(知道最后一个group长度不足)。
Can you solve this real interview question? Reverse Nodes in k-Group - Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list. k is a positive integer and is less than or equal to the length of the linke
更多LeetCode 题解笔记可以访问我的 github。 [TOC] 描述 给出一个链表,每 k 个节点一组进行翻转,并返回翻转后的链表。 k 是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍,那么将最后剩余节点保持原有顺序。 示例: 给定这个链表:1->2->3->4->5 当k = 2 时,应当返回: 2-...
原题链接:https://leetcode.com/problems/reverse-nodes-in-k-group/ 解题思路:这道题是reverse two的升级版,我的代码其实效率很低。首先用了两个标记,一个start指向当前正在reverse的group的第一个,另一个next_start指向下一组的第一个。然后要next_n标记pre要指向的,用cur标记next_n要指向的。一组一组.....