Rotate List

https://oj.leetcode.com/problems/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.

解题思路：

这题目可以说又是一道很ambiguous的题目。这里的k到底是什么意思，很多人都问了这个问题。下面有个比较清晰的解释。来自以下链接。

https://oj.leetcode.com/discuss/2353/what-to-do-when-k-is-greater-than-size-of-list

Let's start with an example.

Given [0,1,2], rotate 1 steps to the right -> [2,0,1].

Given [0,1,2], rotate 2 steps to the right -> [1,2,0].

Given [0,1,2], rotate 3 steps to the right -> [0,1,2].

Given [0,1,2], rotate 4 steps to the right -> [2,0,1].

So, no matter how big K, the number of steps is, the result is always the same as rotating K % n steps to the right.

于是可以看出来，k=k%n，这里n是这个链表里的节点数量。

那么思路就很清楚了，一个快指针，先走k步，然后两个指针一起走，快指针指到结尾停止。再进行指针操作就很清楚了。代码如下。

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode(int x) {

 *         val = x;

 *         next = null;

 *     }

 * }

 */

public class Solution {

    public ListNode rotateRight(ListNode head, int n) {

        if(head == null || n == 0){

            return head;

        }

        ListNode headNode = head;

        ListNode tailNode = head;

        

        int size = 1;

        

        while(tailNode.next != null){

            tailNode = tailNode.next;

            size++;

        }

        

        n = n % size;

        

        if(n == 0){

            return head;

        }

        

        tailNode = head;

        

        for(int i = 0; i < n && tailNode != null; i++){

            tailNode = tailNode.next;

        }

        

        while(tailNode.next != null){

            headNode = headNode.next;

            tailNode = tailNode.next;

        }

        ListNode returNode = headNode.next;

        tailNode.next = head;

        headNode.next = null;

        return returNode;

    }

}