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…

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…

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.…

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-…

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.…

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.…

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…