https://leetcode.com/problems/sort-list/description/
Sort a linked list in O(n log n) time using constant space complexity.
找到中点之后的处理，和 143 "Reorder List" 是一样的 : 然后把tail 传后面去
ListNode tail = mid.next ;
//cut it off
mid.next = null ;
time: o(nlogn) space: o(n): recursive with n stacks
1 public ListNode sortList(ListNode head) {
2 if (head == null || head.next == null) return head ;
3 //merge sort
4 //find the mid,
5 ListNode mid = getMid(head) ;
6 ListNode tail = mid.next ;
7 //cut it off
8 mid.next = null ;
9 //recursive
10 ListNode firstHead = sortList(head) ;
11 ListNode secHead = sortList(tail) ;
12 //merge: same logic as merge two sorted list
13 ListNode newHead = merge(firstHead, secHead);
14 return newHead;
15 }
16
17 private ListNode merge(ListNode firstHead, ListNode secHead) {
18 ListNode dummy = new ListNode(0) ;
19 ListNode curr = dummy ;
20 while (firstHead != null && secHead != null){
21 if (firstHead.val<secHead.val){
22 curr.next = firstHead ;
23 firstHead = firstHead.next ;
24 curr = curr.next ;
25 } else{
26 curr.next = secHead ;
27 secHead = secHead.next ;
28 curr = curr.next ;
29 }
30 }
31 //corner case: either one side left
32 if (firstHead!=null){
33 curr.next = firstHead ;
34 }
35 if (secHead != null){
36 curr.next = secHead ;
37 }
38 return dummy.next ;
39 }
40
41 private ListNode getMid(ListNode head){
42 if (head == null ) return head ;
43 ListNode fast = head , slow = head ;
44 while(fast != null && fast.next != null && fast.next.next != null){
45 fast = fast.next.next ;
46 slow = slow.next ;
47 }
48 return slow ;
49 }