Leetcode: Sort List

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 }