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23. Merge k Sorted Lists

Hard

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

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

import { ListNode } from '../../com_github_leetcode/listnode'

/*
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */
const merge2Lists = (list1: ListNode | null, list2: ListNode | null): ListNode | null => {
    if (!list1 || !list2) {
        return list1 ?? list2
    }
    const tempHead = new ListNode()
    let current = tempHead
    let l1 = list1
    let l2 = list2
    while (l1 || l2) {
        if (!l1) {
            current.next = l2
            break
        }
        if (!l2) {
            current.next = l1
            break
        }
        if (l1.val < l2.val) {
            current.next = l1
            l1 = l1.next
        } else {
            current.next = l2
            l2 = l2.next
        }
        current = current.next
    }
    return tempHead.next
}

const mergeKLists = (lists: Array<ListNode | null>): ListNode | null => {
    while (lists.length > 1) {
        const mergedLists = []
        for (let i = 0; i < lists.length; i += 2) {
            const list1 = lists[i]
            const list2 = lists[i + 1] ?? null
            mergedLists.push(merge2Lists(list1, list2))
        }
        lists = mergedLists
    }
    return lists[0] ?? null
}

export { mergeKLists }

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