725. Split Linked List in Parts

Medium

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.

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

解題

得到節點的總數 ( length ) 後，依照每個 index 需要的節點數量，將相對應長度的區段放入。

Runtime: 0 ms, faster than 100%

Memory Usage: 2.8 MB, less than 20%

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func splitListToParts(head *ListNode, k int) []*ListNode {
    res := make([]*ListNode, k)
    length := 0
    p := head

    for p != nil {
        length++
        p = p.Next
    }

    numNodes := int(math.Floor(float64(length) / float64(k)))
    extraNodes := length % k 
    pre, cur := head, head
    index := 0
    n := numNodes
    if extraNodes > 0 {
        n++
        extraNodes--
    }

    for pre != nil && cur != nil {
        n--
        if n == 0 {
            res[index] = pre
            pre = cur.Next
            cur.Next = nil
            cur = pre

            index++

            n = numNodes
            if extraNodes > 0 {
                n++
                extraNodes--
            }
        } else {
            cur = cur.Next
        }
    }

    return res
}