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 }