Sort List leetcode java

題目：

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

 

題解：

考慮到要求用O(nlogn)的時間複雜度和constant space complexity來sort list，自然而然想到了merge sort方法。同時我們還已經做過了merge k sorted list和merge 2 sorted list。這樣這個問題就比較容易了。

不過這道題要找linkedlist中點，那當然就要用最經典的faster和slower方法，faster速度是slower的兩倍，當faster到鏈尾時，slower就是中點，slower的next是下一半的開始點。

解決了這些問題，題目就很容易解出了。

 程式碼如下：

 1     public ListNode sortList(ListNode head) {
 2         if(head == null|| head.next == null)
 3             return head;
 4         ListNode slow = head, fast = head, firsthalf = head;
 5         while(fast.next!=null&&fast.next.next!=null){
 6             slow = slow.next;
 7             fast = fast.next.next;
 8         }
 9         ListNode secondhalf = slow.next;
10         slow.next = null;
11         
12         ListNode leftlist = null, rightlist =null;
13         if(firsthalf!=secondhalf){
14             leftlist = sortList(firsthalf);
15             rightlist = sortList(secondhalf);
16         }
17         return mergeTwoLists(leftlist, rightlist);
18     }
19     
20     public ListNode mergeTwoLists(ListNode leftlist, ListNode rightlist){
21         if(rightlist == null)
22             return leftlist;
23         if(leftlist == null)
24             return rightlist;
25         
26         ListNode fakehead = new ListNode(-1);
27         ListNode ptr = fakehead;
28         while(rightlist!=null&&leftlist!=null){
29             if(rightlist.val<leftlist.val){
30                 ptr.next = rightlist;
31                 ptr = ptr.next;
32                 rightlist = rightlist.next;
33             }else{
34                 ptr.next = leftlist;
35                 ptr = ptr.next;
36                 leftlist = leftlist.next;
37             }
38         }
39         
40         if(rightlist!=null)
41             ptr.next = rightlist;
42         if(leftlist!=null)
43             ptr.next = leftlist;
44         
45         return fakehead.next;
46     }