Intuition

The initial thought is to split the linked list into k parts, ensuring each part is as evenly distributed as possible. If the number of nodes in the list isn’t evenly divisible by k, the first few parts should contain one additional node.

Approach

1. Calculate Length: Begin by calculating the total length of the linked list.
2. Determine Part Sizes: Divide the length by k to determine the base size of each part. The remainder will help in distributing extra nodes to the first few parts.
3. Split the List: Iterate through the linked list, assigning the appropriate number of nodes to each part while making sure to sever the connections between parts properly.

Complexity

• Time complexity:
  The time complexity is $O(n)$, where n is the total number of nodes in the linked list. This accounts for traversing the entire list to determine its length and then splitting the list.

• Space complexity:
  The space complexity is $O(k)$ for storing the resulting array of linked list parts.

Code

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 * int val;
 * ListNode next;
 * ListNode() {}
 * ListNode(int val) { this.val = val; }
 * ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode[] splitListToParts(ListNode head, int k) {
        int N = getLength(head);

        int lengthPerPart = N / k;
        int leftOver = N % k;

        ListNode[] result = new ListNode[k];
        ListNode current = head;

        for (int i = 0; i < k; i++) {
            result[i] = current;
            int partSize = lengthPerPart + (i < leftOver ? 1 : 0);

            for (int j = 0; j < partSize - 1; j++) {
                if (current != null) {
                    current = current.next;
                }
            }

            if (current != null) {
                ListNode next = current.next;
                current.next = null;
                current = next;
            }
        }

        return result;
    }

    private int getLength(ListNode head) {
        int count = 0;
        ListNode current = head;
        while (current != null) {
            count++;
            current = current.next;
        }
        return count;
    }
}

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func splitListToParts(head *ListNode, k int) []*ListNode {
    current, n := head, 0
    for current != nil {
        n++
        current = current.Next
    }
    remain, size := n % k, n / k
    ans := make([]*ListNode, k)
    current = head
    for index := 0; index < k && current != nil; index++ {
        dummy := &ListNode{}
        first := dummy
        partSize := size
        if remain > 0 {
            partSize++
            remain--
        }
        for i := 0;  i < partSize; i++ {
            first.Next = &ListNode{Val: current.Val}
            first = first.Next
            current = current.Next
        }
        ans[index] = dummy.Next
    }
    return ans
}