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 }

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