Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts". The length of each part should be as equal as possible: no two parts should have a size differing by more than 1.…

［抄题］: Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts". The length of each part should be as equal as possible: no two parts should have a size differing by more t…

https://leetcode.com/problems/split-linked-list-in-parts/ Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts". The length of each part should be as equal as possible:…

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts". The length of each part should be as equal as possible: no two parts should have a size differing by more than 1.…

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts". The length of each part should be as equal as possible: no two parts should have a size differing by more than 1.…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/split-linked-list-in-parts/description/ 题目描述 Given a (singly) linked list with head node root, write a function to split the linked list i…

▶ 将一个单链表拆分为长度尽量接近的 k 段 ● 自己的代码,12 ms ■ 记链表长度为 count,目标段数为 k,quo = count / k,mod = count % k,part = mod * (quo + 1) ■ 前半截(长半截)共有 mod 组,每组 quo + 1 个元素,共 mod * (quo + 1) 个元素,这是 part 的由来:后半截(长半截)共有 k - mod 组,每组 quo 个元素,共 quo * (k - mod) 个元素 ■ 当 i < part…

题目描述:给定一个单链表,写一个函数把它分成k个单链表.分割成的k个单链表中,两两之间长度差不超过1,允许为空.分成的k个链表中,顺序要和原先的保持一致,比如说每个单链表有3个结点,则第一个单链表的结点为输入链表的前三个结点,依次类推. 思路: 第一次遍历单链表,求出链表的长度length: 求出平均分成的k个链表中,每个的结点avg,以及还多余的结点rem: 第二次遍历输入链表,如果达到avg,且rem存在值,则把本次遍历的结果赋值给结果数组: # Definition for singly-…

题意:将原链表分隔成k个链表,要求所有分隔的链表长度差异至多为1,且前面的链表长度必须大于等于后面的链表长度. 分析: (1)首先计算链表总长len (2)根据len得到分隔的链表长度要么为size,要么为size+1,由于前面的链表长度必须大于等于后面的链表长度,因此,前mod个分隔的链表长度为size+1,其他分隔的链表长度为size (3) vector<ListNode*> ans(k)----k个ListNode*,每个元素都初始化为NULL:对于原链表为NULL,或是len<…

思路很简单  按时链表的题做起来很容易犯小错误,思维要缜密 还要多练习啊 做之前最好画算法框图 public ListNode[] splitListToParts(ListNode root, int k) { ListNode[] res = new ListNode[k]; int count = 0; ListNode temp = root; //计算长度 while (temp!=null) { count++; temp = temp.next; } //计算链表长度 int si…

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts". The length of each part should be as equal as possible: no two parts should have a size differing by more than 1.…

Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts". The length of each part should be as equal as possible: no two parts should have a size differing by more than 1.…

1.题目描述 2.题目分析 主要是理解题意,将每个子链表应该分得的节点个数计算清楚.利用除数和余数的方法进行计算. 3.代码 vector<ListNode*> splitListToParts(ListNode* root, int k) { vector<ListNode*> res(k, NULL); if (root == NULL) { return res; } ; ListNode *p = root; while (p != NULL) { listlen++; p…

突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…

645. Set Mismatch The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another nu…

All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 如果各位看官们,大神们发现了任何错误,或是代码无法通过OJ,或是有更好的解法,或是有任何疑问,意见和建议的话,请一定要在对应的帖子下面评论区留言告知博主啊(如果不方便注册博客园的话,可以下载下文提到的APP,在Feedback中给博主发邮件交流哈),同时也请大家踊跃地,大量地,盲目地提供各个题目的follow up一起讨论哈,多谢多谢,祝大家刷得愉快…

1.发现两个链表的交点 160.两个链表的交集(容易) Leetcode /力扣 public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { ListNode curA=headA; ListNode curB=headB; while(curA!=curB){ if(curA==null){ curA=headB; }else{ curA=curA.next; } i…

328. 奇偶链表 328. Odd Even Linked List 题目描述 给定一个单链表,把所有的奇数节点和偶数节点分别排在一起.请注意,这里的奇数节点和偶数节点指的是节点编号的奇偶性,而不是节点的值的奇偶性. 请尝试使用原地算法完成.你的算法的空间复杂度应为 O(1),时间复杂度应为 O(nodes),nodes 为节点总数. LeetCode328. Odd Even Linked List中等 示例 1: 输入: 1->2->3->4->5->NULL 输出:…

[2]Add Two Numbers (2018年11月30日,第一次review,ko) 两个链表,代表两个整数的逆序,返回一个链表,代表两个整数相加和的逆序. Example: Input: ( -> -> ) + ( -> -> ) Output: -> -> Explanation: + = . /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *…

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 把自己刷过的所有题目做一个整理,并且用简洁的语言概括了一下思路,汇总成了一个表格. 题目的排列顺序是按照先Easy再Medium再Hard排列的,暂时还没有把题目全部整理完成.后序我会把刷过的所有的题目都整理到这个文档里. 题目 难度 解法 题目地址 566. Reshape the Matrix Easy 变长数组,求余法,维护行列计算在新的数组中的位置 https://blog.c…

When file is too large, we will compress it and split it into several parts. Now Let me show you how to combine them in Linux command line. step 1: Install rar command sudo apt-get install rar sudo apt-get install p7zip-full step 2: Merge rar files r…

Total Commander 8.52 Beta 1http://www.ghisler.com/852_b1.php 10.08.15 Release Total Commander 8.52 beta 1 (32/64) 05.08.15 Fixed: Windows 10: Loading drive buttonbar hanging on some devices (e.g. Surface Pro 3) when SD-Card was in internal card reade…

Introduction the naive "scull" 首先.什么是scull? scull (Simple Character Utility for Loading Localities). scull is a char driver that acts on a memory area as though it were a device. 和第一个C程序Hello world一样.他什么都不能干,却能非常好的阐释怎么一步步进阶的去写驱动 blog的最后,我会给出这对于s…

Modules environmentDescription This is a system that allows you to easily change between different versions of compilers and other software, no matter what shell you use, without having to set a lot of environment variables by hand each time. The aut…

转自:https://www.linuxtv.org/downloads/legacy/video4linux/API/V4L2_API/spec-single/v4l2.html Video for Linux Two API Specification Revision 2.6.32 Michael H Schimek <mschimek@gmx.at> Bill Dirks Original author of the V4L2 API and documentation. Hans V…

awesome-android Introduction android libs from github System requirements Android Notice If the lib is no longer being maintained,please do not add it here. How To Contribute Step 1. Add a Item as follows: **Library Name**[one space]Short Description…

由于Linux内核版本更新的原因,LDD3(v2.6.10)提供的源码无法直接使用,下面是本人编译scull源码时出现的一些问题及解决方法.编译环境:Ubuntu 10.04 LTS(kernel version 2.6.32-33) 编译错误: make -C /lib/modules/-.el6.i686/build M=/mnt/HappyStudy/MyDesigner/Linux/LDD3/examples/scull LDDINC=/mnt/HappyStudy/MyDesigner…

Linux对内核态内存分配请求与用户态内存分配请求处理上分别对待 Linux本身信任自己,因此Linux内核请求分配多少内存,就会马上分配相同数量的内存出来. 但内核本身不相信应用程序,而且通常应用程序分配了一段内存,其实只是预定,并不是马上就去访问.由于应用程序的数目比较多,那么这部分只分配了但是没有立即访问的内存就占了很大的比例. 1. 因此,内核通过Page Fault exception handler来延迟(Defer)对应用程序申请的内存进行分配操作. 2. 用户态的应用程序分配内存…