LeetCode解题报告:Reorder List

Reorder List

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes‘ values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

思路：

1.利用快慢两个指针将链表一分为二；

2.针对第二个子链表求倒序；

3.利用merge思想将两个子链表合并。

代码：

 1 public class ReorderSingleList {
 2     /**
 3      * Definition for singly-linked list. class ListNode { int val; ListNode
 4      * next; ListNode(int x) { val = x; next = null; } }
 5      */
 6     public void reorderList(ListNode head) {
 7         if (head == null || (head.next == null)) {
 8             return;
 9         }
10         // fast and slow point find the mid position.
11         ListNode fast = head;
12         ListNode slow = head;
13         while ((fast != null) && (fast.next != null)) {
14             fast = fast.next.next;
15             slow = slow.next;
16         }
17
18         // reverse the last second list.
19         ListNode headnode = new ListNode(-1);
20         headnode.next = slow;
21         ListNode temp = headnode.next;
22         while (temp.next != null) {
23             ListNode insert = temp.next;
24             ListNode currNext = insert.next;
25             insert.next = headnode.next;
26             headnode.next = insert;
27             temp.next = currNext;
28         }
29
30         // merge insert
31         ListNode firstcur = head;
32         ListNode secondcur = headnode.next;
33         while (firstcur != slow && (secondcur != slow)) {// at first,I make a mistake in here;
34             ListNode firstnex = firstcur.next;
35             ListNode secondnex = secondcur.next;
36             firstcur.next = secondcur;
37             secondcur.next = firstnex;
38             firstcur = firstnex;
39             secondcur = secondnex;
40         }
41     }
42
43     public static void main(String[] args) {
44         ListNode t5 = new ListNode(5);
45         ListNode t4 = new ListNode(4, t5);
46         ListNode t3 = new ListNode(3, t4);
47         ListNode t2 = new ListNode(2, t3);
48         ListNode t1 = new ListNode(1, t2);
49         new ReorderSingleList().reorderList(t1);
50         System.out.println(t1);
51     }
52
53 }

LeetCode解题报告:Reorder List