执行用时 :4 ms, 在所有 Java 提交中击败了98.35% 的用户
内存消耗 :38.9 MB, 在所有 Java 提交中击败了94.11%的用户
原本是“逐一两两合并链表”，用时118ms，换成假分治后只用4ms，震惊！代码其实就是把遍历换成了队列……
关键代码变化：
“分治”：
1 static public ListNode mergeKLists(ListNode[] lists) {
2 if (lists == null || lists.length == 0) return null;
3 Queue<ListNode> q = new ArrayDeque<>();
4 for (ListNode n : lists) if (n != null) q.add(n);
5 while (q.size() > 1) {
6 q.add(merge2(q.poll(), q.poll()));
7 }
8 return q.poll();
9 }
逐一两两合并链表：
1 public ListNode mergeKLists(ListNode[] lists) {
2 if (lists.length==0)return null;
3 ListNode head = lists[0];
4 for (int i = 1; i < lists.length; i++) head = merge2(head, lists[i]);
5 return head;
6 }
有毒……
1 import java.util.ArrayDeque;
2 import java.util.Queue;
3
4 public class Solution {
5 static public ListNode mergeKLists(ListNode[] lists) {
6 if (lists == null || lists.length == 0) return null;
7 Queue<ListNode> q = new ArrayDeque<>();
8 for (ListNode n : lists) if (n != null) q.add(n);
9 while (q.size() > 1) {
10 q.add(merge2(q.poll(), q.poll()));
11 }
12 return q.poll();
13 }
14
15 static ListNode merge2(ListNode l1, ListNode l2) {
16 if (l1 == null) return l2;
17 if (l2 == null) return l1;
18
19 if (l1.val > l2.val) {
20 ListNode t = l1;
21 l1 = l2;
22 l2 = t;
23 }
24
25 ListNode head = l1;
26 while (l1 != null && l2 != null) {
27 while (l1.next != null && l1.next.val < l2.val) l1 = l1.next;
28 if (l1.next == null) {
29 l1.next = l2;
30 return head;
31 }
32 ListNode next = l1.next;
33 l1.next = l2;
34 while (l2.next != null && l2.next.val < next.val) l2 = l2.next;
35 ListNode t = l2;
36 l2 = l2.next;
37 t.next = next;
38 l1 = next;
39 }
40 return head;
41 }
42 }