725. Split Linked List in Parts #

Problem #

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].

Problem Summary #

Divide the linked list into K parts，requiring that the lengths of these K parts differ pairwise by no more than 1 as much as possible，and that the lengths are as equal as possible。

Solution Approach #

Divide the length of the linked list by K，the result is the final length n of each group。Take the length of the linked list modulo K，the result m obtained represents that the length of each of the first m groups of the linked list is n + 1 。This is equivalent to distributing all the extra parts into the first m groups of the linked list。In the end，the first m groups of the linked list have length n + 1，and the remaining K - m groups of the linked list have length n。

Note that when the length is less than K，use nil for padding。

Code #

package leetcode

import "fmt"

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func splitListToParts(root *ListNode, k int) []*ListNode {
	res := make([]*ListNode, 0)
	if root == nil {
		for i := 0; i < k; i++ {
			res = append(res, nil)
		}
		return res
	}
	length := getLength(root)
	splitNum := length / k
	lengNum := length % k
	cur, head := root, root
	var pre *ListNode
	fmt.Printf("total length %v, split into %v groups, first %v groups have length %v, remaining %v groups，each group %v\n", length, k, lengNum, splitNum+1, k-lengNum, splitNum)
	if splitNum == 0 {
		for i := 0; i < k; i++ {
			if cur != nil {
				pre = cur.Next
				cur.Next = nil
				res = append(res, cur)
				cur = pre
			} else {
				res = append(res, nil)
			}
		}
		return res
	}
	for i := 0; i < lengNum; i++ {
		for j := 0; j < splitNum; j++ {
			cur = cur.Next
		}
		fmt.Printf("0 just came out head = %v cur = %v pre = %v\n", head, cur, head)
		pre = cur.Next
		cur.Next = nil
		res = append(res, head)
		head = pre
		cur = pre
		fmt.Printf("0 head = %v cur = %v pre = %v\n", head, cur, head)
	}
	for i := 0; i < k-lengNum; i++ {
		for j := 0; j < splitNum-1; j++ {
			cur = cur.Next
		}
		fmt.Printf("1 just came out head = %v cur = %v pre = %v\n", head, cur, head)
		pre = cur.Next
		cur.Next = nil
		res = append(res, head)
		head = pre
		cur = pre
	}
	return res
}

func getLength(l *ListNode) int {
	count := 0
	cur := l
	for cur != nil {
		count++
		cur = cur.Next
	}
	return count
}