mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-05 08:27:30 +08:00
58 lines
2.1 KiB
Markdown
58 lines
2.1 KiB
Markdown
# [725. Split Linked List in Parts](https://leetcode.com/problems/split-linked-list-in-parts/)
|
||
|
||
## 题目
|
||
|
||
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:
|
||
|
||
```c
|
||
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:
|
||
|
||
```c
|
||
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].
|
||
|
||
|
||
|
||
## 题目大意
|
||
|
||
把链表分成 K 个部分，要求这 K 个部分尽量两两长度相差不超过 1，并且长度尽量相同。
|
||
|
||
## 解题思路
|
||
|
||
把链表长度对 K 进行除法，结果就是最终每组的长度 n。把链表长度对 K 进行取余操作，得到的结果 m，代表前 m 组链表长度为 n + 1 。相当于把多出来的部分都分摊到前面 m 组链表中了。最终链表是前 m 组长度为 n + 1，后 K - m 组链表长度是 n。
|
||
|
||
注意长度不足 K 的时候要用 nil 进行填充。
|
||
|
||
|
||
|