LeetCode OJ - Reorder List

这道题参考了Discuss 中一个前辈的思路：把输入的链表分成前后两半，将后半链表reverse，最后merge成目标链表。

 

 1 /**
 2      * reorder the list in-place without altering the nodes' value
 3      * 参考了讨论中前辈提供的一种思路O(N)时间复杂度、O(1)空间复杂度
 4      * @param head
 5      */
 6     public void reorderList(ListNode head){
 7         if(head == null || head.next == null)
 8             return ;
 9         int len = 0;
10         ListNode curr = head,temp = null;
11         //caculate the length of the list
12         while(curr!=null)
13         {
14             len ++ ;
15             curr = curr.next;
16         }
17         
18         int i = 0;
19         temp = null; // after the while, temp is the end of the first half List;
20         curr = head;// after the while, curr is the head of the second half List
21         while(i++<(len-len/2))
22         {
23             temp = curr;
24             curr = curr.next; 
25         }
26         temp.next = null;// the next of the new end should be null
27         
28         //reverse the second half List
29         ListNode shead = curr;
30         ListNode po = curr.next;
31         shead.next = null;
32         while(po!=null)
33         {
34             ListNode t = po.next;
35             po.next = shead;
36             shead = po;
37             po = t;
38         }
39         
40         //merge the first half and the second reversed half
41         po = head;
42         while(shead !=null){
43             ListNode f1 = po.next;
44             ListNode f2 = shead.next;
45             po.next = shead;
46             shead.next = f1;
47             po = f1;
48             shead = f2;        
49         }
50     }

 

转载于:https://www.cnblogs.com/echoht/p/3687115.html