[LeetCode] 23. Merge k Sorted Lists（合并 k 个有序链表）

• Difficulty: Hard
• Related Topics: Linked List, Divide and Conquer, Heap
• Link: https://leetcode.com/problems/merge-k-sorted-lists/

Description

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.
给定包含 k 个单链表的数组 lists，每个链表都已升序排序。

Merge all the linked-lists into one sorted linked-list and return it.
合并所有链表为一个链表并返回之。

Examples

Example 1

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2

Input: lists = []
Output: []

Example 3

Input: lists = [[]]
Output: []

Constraints

• k == lists.length
• 0 <= k <= 10^4
• 0 <= lists[i].length <= 500
• -10^4 <= lists[i][j] <= 10^4
• lists[i] is sorted in ascending order.
• The sum of lists[i].length won't exceed 10^4.

Solution

这题的解题思路与归并排序十分相似，我们只要先把 lists 的左半部份合并，再将其右半部份合并，最后将两个合并的结果合并即可，代码如下：

class Solution {
    fun mergeKLists(lists: Array<ListNode?>): ListNode? {
        if (lists.isEmpty()) {
            return null
        }
        if (lists.size == 1) {
            return lists[0]
        }
        val mid = lists.size / 2
        val leftResult = mergeKLists(lists.sliceArray(0 until mid))
        val rightResult = mergeKLists(lists.sliceArray(mid until lists.size))
        return mergeResult(leftResult, rightResult)
    }

    private fun mergeResult(left: ListNode?, right: ListNode?): ListNode? {
        val dummy = ListNode(-1)
        var p = left
        var q = right
        var r: ListNode? = dummy

        while (p != null && q != null) {
            if (p.`val` < q.`val`) {
                r?.next = p
                p = p.next
            } else {
                r?.next = q
                q = q.next
            }
            r = r?.next
        }

        if (p != null) {
            r?.next = p
        }
        if (q != null) {
            r?.next = q
        }

        return dummy.next
    }
}