Leetcode: Sort List

最新推荐文章于 2017-05-23 12:21:29 发布

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最新推荐文章于 2017-05-23 12:21:29 发布

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原文链接：http://www.cnblogs.com/EdwardLiu/p/3978383.html

本文介绍了一种使用常数空间复杂度在O(nlogn)时间内对链表进行排序的方法。通过采用归并排序的思想，利用runner technique将链表分成两半，递归地排序每一半，然后合并两个有序链表。

Sort a linked list in O(n log n) time using constant space complexity.

记得Insert Sort List, 那个复杂度是O(N^2)的，这里要求O（nlogn），所以想到merge sort, 需要用到Merge Two Sorted List的方法（我写的merge函数）

第二遍代码：所以mergesort不好写在于它是一个递归里嵌套另一个递归，第一个递归不停地用runner technique把list分两段，直到每一段是一个或0个节点返回该点，这时再调用merge Two sorted List递归把两段整合起来，返回它们的首节点

 1 public class Solution {
 2     public ListNode sortList(ListNode head) {
 3         if (head == null || head.next == null) return head;
 4         return mergesort(head);
 5     }
 6     
 7     public ListNode mergesort(ListNode head) {
 8         if (head == null || head.next == null) return head;
 9         ListNode dummy = new ListNode(-1);
10         dummy.next = head;
11         ListNode walker = dummy;
12         ListNode runner = dummy;
13         while (runner!=null && runner.next!=null) {
14             runner = runner.next.next;
15             walker = walker.next;
16         }
17         ListNode head1 = dummy.next;
18         ListNode head2 = walker.next;
19         walker.next = null;
20         return merge(mergesort(head1), mergesort(head2));
21     }
22     
23     public ListNode merge(ListNode head1, ListNode head2) {
24         if (head1 == null) return head2;
25         if (head2 == null) return head1;
26         ListNode dummy = new ListNode(-1);
27         dummy.next = head1;
28         ListNode pre = dummy;
29         while (head1 != null && head2 != null) {
30             if (head1.val <= head2.val) {
31                 head1 = head1.next;
32             }
33             else {
34                 ListNode next = head2.next;
35                 head2.next = pre.next;
36                 pre.next = head2;
37                 head2 = next;
38             }
39             pre = pre.next;
40         }
41         if (head2 != null) {
42             pre.next = head2;
43         }
44         return dummy.next;
45     }
46 }

转载于:https://www.cnblogs.com/EdwardLiu/p/3978383.html