[LeetCode] Split Linked List in Parts 分割链表成部分

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. This may lead to some parts being null.

The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than 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 [ [1], [2], [3], [4], null ]

Example 1:

Input:
root = [1, 2, 3], k = 5
Output: [[1],[2],[3],[],[]]
Explanation:
The input and each element of the output are ListNodes, not arrays.
For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.
The first element output[0] has output[0].val = 1, output[0].next = null.
The last element output[4] is null, but it‘s string representation as a ListNode is [].

Example 2:

Input:
root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
Explanation:
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.

Note:

  • The length of root will be in the range [0, 1000].
  • Each value of a node in the input will be an integer in the range [0, 999].
  • k will be an integer in the range [1, 50].

s

时间: 2024-10-29 16:12:06

[LeetCode] Split Linked List in Parts 分割链表成部分的相关文章

LeetCode 725. Split Linked List in Parts (分裂链表)

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.

725. Split Linked List in Parts 拆分链表

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.

725. Split Linked List in Parts把链表分成长度不超过1的若干部分

[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

[LeetCode] Split Array With Same Average 分割数组成相同平均值的小数组

In a given integer array A, we must move every element of A to either list B or list C. (B and C initially start empty.) Return true if and only if after such a move, it is possible that the average value of B is equal to the average value of C, and

[LeetCode] Split Array with Equal Sum 分割数组成和相同的子数组

Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions: 0 < i, i + 1 < j, j + 1 < k < n - 1 Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be

LC 725. 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: no two parts should have a size differing by more than 1.

#Leetcode# 725. Split Linked List in Parts

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:

[Swift]LeetCode725. 分隔链表 | 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: no two parts should have a size differing by more than 1.

LeetCode OJ:Partition List(分割链表)

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. For example,Given 1->4->3->2