LeetCode 23. Merge k Sorted Lists (javascript)

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.

Idea:

Divide and Conquer

Solution:

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode[]} lists
 * @return {ListNode}
 */
var mergeKLists = function(lists) {
    return merge(lists, 0, lists.length - 1);
};

function merge(lists, l , r) {
    if (lists.length === 0) return null;
    if (l === r) return lists[l];
    let mid = Math.floor((l + r) / 2);
    let l1 = merge(lists, l, mid);
    let l2 = merge(lists, mid + 1, r);
    return mergeTwoLists(l1, l2);
}

var mergeTwoLists = function(l1, l2) {
    // If one of the list is empty, return the other one.
      if (l1 === null) {
        return l2;
      }
      if (l2 === null) {
        return l1;
      }
    
    // The smaller one becomes the head.
    if (l1.val < l2.val) {
        l1.next = mergeTwoLists(l1.next, l2);
        return l1;
    } else {
        l2.next = mergeTwoLists(l1, l2.next);
        return l2;
    }
};