Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

最开始想直接用双指针做就行了。 但是发现n竟然可以比length长。

 1 /**

 2  * Definition for singly-linked list.

 3  * public class ListNode {

 4  *     int val;

 5  *     ListNode next;

 6  *     ListNode(int x) {

 7  *         val = x;

 8  *         next = null;

 9  *     }

10  * }

11  */

12 public class Solution {

13     public ListNode rotateRight(ListNode head, int n) {

14         // Note: The Solution object is instantiated only once and is reused by each test case.

15         ListNode fast = head;

16         ListNode cur = head;

17         if(n == 0) return head;

18         if(head == null)return null;

19         for(int i = 0; i < n; i ++){

20             if(fast == null) return head;

21             fast = fast.next;

22         }

23         if(fast == null) return head;

24         while(fast.next != null){

25             fast = fast.next;

26             cur = cur.next;

27         }

28         ListNode hd = cur.next;

29         cur.next = null;

30         fast.next = head;

31         return hd;

32     }

33 }

Solution：

首先从head开始跑，直到最后一个节点，这时可以得出链表长度len。然后将尾指针指向头指针，将整个圈连起来，接着往前跑len – k%len，从这里断开，就是要求的结果了。

 1 /**

 2  * Definition for singly-linked list.

 3  * public class ListNode {

 4  *     int val;

 5  *     ListNode next;

 6  *     ListNode(int x) {

 7  *         val = x;

 8  *         next = null;

 9  *     }

10  * }

11  */

12 public class Solution {

13     public ListNode rotateRight(ListNode head, int n) {

14         // Note: The Solution object is instantiated only once and is reused by each test case.

15        ListNode cur = head;

16        int len = 1;

17        if(head == null) return null;

18        while(cur.next != null){

19            cur = cur.next;

20            len ++;

21        }

22        cur.next = head;

23        int t= len - n % len;

24        for(int i = 0; i < t; i ++){

25            cur = cur.next;

26        }

27        ListNode hd = cur.next;

28        cur.next = null;

29        return hd;

30     }

31 }