LeetCode 第23题 合并K个排序链表

/*

23. 合并K个排序链表

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入:[  1->4->5,  1->3->4,  2->6]输出: 1->1->2->3->4->4->5->6 */

/** * Definition for singly-linked list.  public class ListNode { int val; ListNode next; ListNode(int * x) { val = x; } } */

/*思路1: 分治法，归并排序思路2: 优先队列*/

 1 class Solution23 {
 2
 3   /*
 4   分治法，归并排序.
 5    */
 6   public ListNode mergeKLists(ListNode[] lists) {
 7     if (lists == null || lists.length == 0) {
 8       return null;
 9     }
10     return sort(lists, 0, lists.length - 1);
11   }
12
13   ListNode sort(ListNode[] list, int left, int right) {
14     if (left < right) {
15       int mid = (left + right) >> 1;
16       ListNode le = sort(list, left, mid);
17       ListNode ri = sort(list, mid + 1, right);
18       return merge(le, ri);
19     }
20     return list[left];
21   }
22
23   ListNode merge(ListNode le, ListNode ri) {
24     if (le == null) {
25       return ri;
26     }
27     if (ri == null) {
28       return le;
29     }
30     if (le.val < ri.val) {
31       le.next = merge(le.next, ri);
32       return le;
33     } else {
34       ri.next = merge(le, ri.next);
35       return ri;
36     }
37   }
38
39   /*
40     优先队列式.
41    */
42   public ListNode mergeKLists2(ListNode[] lists) {
43
44     if (lists == null || lists.length == 0) {
45       return null;
46     }
47     if (lists.length == 1) {
48       return lists[0];
49     }
50     PriorityQueue<ListNode> queue = new PriorityQueue<>(Comparator.comparingInt(o -> o.val));
51     for (ListNode list : lists) {
52       if (list != null) {
53         queue.add(list);
54       }
55     }
56     ListNode dummy = new ListNode(0);
57     ListNode curr = dummy;
58     while (!queue.isEmpty()) {
59       curr.next = queue.poll();
60       curr = curr.next;
61       if (curr.next != null) {
62         queue.add(curr.next);
63       }
64     }
65     return dummy.next;
66   }
67 }

原文地址：https://www.cnblogs.com/rainbow-/p/10296459.html