Leetcode: Sort List

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

 1 /**
 2  * Definition for singly-linked list.
 3  * class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode sortList(ListNode head) {
14         if (head == null) {
15             return null;
16         }
17         return mergesort(head);
18     }
19     
20     public ListNode mergesort(ListNode head) {
21         if (head.next == null) return head;
22         ListNode dummy = new ListNode(0);
23         dummy.next = head;
24         ListNode current = dummy;
25         ListNode runner = dummy;
26         while (runner != null && runner.next != null) {
27             runner = runner.next.next;
28             current = current.next;
29         }
30         ListNode newhead = current.next;
31         current.next = null;
32         ListNode head1 = mergesort(head);
33         ListNode head2 = mergesort(newhead);
34         return merge(head1, head2);
35     }
36     
37     public ListNode merge(ListNode l1, ListNode l2) {
38         ListNode dummyNode = new ListNode(0);
39         ListNode currentNode = dummyNode;
40         
41         while (l1!=null || l2!=null) {
42             if (l1!=null && l2!=null) {
43                 if (l1.val > l2.val) {
44                     currentNode.next = l2;
45                     l2 = l2.next;
46                 }
47                 else {
48                     currentNode.next = l1;
49                     l1 = l1.next;
50                 }
51             }
52             else {
53                 if (l1 != null) {
54                     currentNode.next = l1;
55                     l1 = l1.next;
56                 }
57                 else {
58                     currentNode.next = l2;
59                     l2 = l2.next;
60                 }
61             }
62             currentNode = currentNode.next;
63         }
64         return dummyNode.next;
65     }
66 }