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Split Linked List in Parts

Problem Statement​

Given the head of a singly linked list and an integer k, 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 one. This may lead to some parts being null.

The parts should be in the order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal to parts occurring later.

Return an array of the k parts.

Leetcode Link

Example 1:​

Input: head = [1,2,3], k = 5
Output: [[1],[2],[3],[],[]]
Explanation:
The first element output[0] has output[0].val = 1, output[0].next = null.
The last element output[4] is null, but its string representation as a ListNode is [].

Example 2:​

Input: head = [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.

Constraints:​

  • The number of nodes in the list is in the range [0, 1000].
  • 0 <= Node.val <= 1000
  • 1 <= k <= 50

Code​

Python Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def splitListToParts(self, root, k):
        # Count the length of the linked list
        curr, length = root, 0
        while curr:
            curr, length = curr.next, length + 1
        # Determine the length of each chunk
        chunk_size, longer_chunks = length // k, length % k
        res = [chunk_size + 1] * longer_chunks + [chunk_size] * (k - longer_chunks)
        # Split up the list
        prev, curr = None, root
        for index, num in enumerate(res):
            if prev:
                prev.next = None
            res[index] = curr
            for i in range(num):
                prev, curr = curr, curr.next
        return res