LeetCode题解之Split Linked List in Parts 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 1 vector<ListNode*> splitListToParts(ListNode* root, int k) { 2 vector<ListNode*> res(k, N
The parts should be in order of occurrence in the input list, andparts occurring earlier should always have a size greaterthan or equal parts occurring later. Return a List of ListNode's representing the linked list parts that are formed. Examples 1->2->3->4, k = 5 // 5 equal parts...
Given a (singly) linked list with head noderoot, 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. This may lead to some parts being nul...
725 Split Linked List in Parts.py4.11 KB 一键复制编辑原始数据按行查看历史 Loading... 跳转 举报 举报成功 我们将于2个工作日内通过站内信反馈结果给你! 请认真填写举报原因,尽可能描述详细。 举报类型 请选择举报类型 举报原因 取消 发送 误判申诉 ...
【Leetcode】725. Split Linked List in Parts 1 由于每一个chunk结束的时候,需要指向None,所以需要一个pre指针来记录走到最后的位置
题目链接[https://leetcode-cn.com/problems/split-linked-list-in-parts/]tag: Medium Linked qu...
leetcode 725 Split Linked List in Parts 1. len/k=a...b,共k个part,每个part有a个节点,余下的b的节点,从第一个节点开始,每个节点多加一个。 classSolution {public: vector<ListNode*> splitListToParts(ListNode* root,intk) { vector<ListNode*>vec(k,nullptr);if(!k)returnvec;intlen=0;...
The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts. 题目标签:Linked List 题目给了我们一个链表,还给了我们k,让我们把链表分成 k 个部分,使得每一个部分尽可能相等。
package LeetCode_725 /** * 725. Split Linked List in Parts * https://leetcode.com/problems/split-linked-list-in-parts/description/ * * Given a (singly
链接:https://leetcode-cn.com/problems/split-linked-list-in-parts 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 题意不难理解,给一个 linked list,请你按照规则分割成若干个子 linked list,以 ListNode[] 的形式输出。分割的要求是需要把 input 分成 K 个组,任何两组 node 之间的...