[LeetCode] 23. Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

合并K个排序链表。题目即是题意。这个题的最优解应该是用priority queue解决但是因为JS实现PQ太过麻烦，所以JS实现我给出次优解，思路是merge sort。

时间O(nlogk) - k是链表的数量

空间O(n)

JavaScript实现

 1 /**
 2  * @param {ListNode[]} lists
 3  * @return {ListNode}
 4  */
 5 var mergeKLists = function (lists) {
 6     return divide(lists, 0, lists.length - 1);
 7 };
 8 
 9 var divide = function (lists, start, end) {
10     if (start === end) {
11         return lists[start];
12     } else if (start < end) {
13         const mid = parseInt(start + (end - start) / 2);
14         const left = divide(lists, start, mid);
15         const right = divide(lists, mid + 1, end);
16         return merge(left, right);
17     } else {
18         return null;
19     }
20 }
21 
22 var merge = function (left, right) {
23     if (!left) {
24         return right;
25     } else if (!right) {
26         return left;
27     } else if (left.val < right.val) {
28         left.next = merge(left.next, right);
29         return left;
30     } else {
31         right.next = merge(left, right.next);
32         return right;
33     }
34 }

Java实现 - priority queue

时间O(nlogk) - k是链表的数量

空间O(n)

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode() {}
 7  *     ListNode(int val) { this.val = val; }
 8  *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 9  * }
10  */
11 class Solution {
12     public ListNode mergeKLists(ListNode[] lists) {
13         // corner case
14         if (lists == null || lists.length == 0) {
15             return null;
16         }
17 
18         // normal case
19         PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, (a, b) -> a.val - b.val);
20         ListNode dummy = new ListNode(0);
21         ListNode cur = dummy;
22         for (ListNode list : lists) {
23             // 注意sublist有可能是空的
24             if (list != null) {
25                 queue.add(list);
26             }
27         }
28 
29         while (!queue.isEmpty()) {
30             cur.next = queue.poll();
31             cur = cur.next;
32             if (cur.next != null) {
33                 queue.add(cur.next);
34             }
35         }
36         return dummy.next;
37     }
38 }