61. Rotate List

QUESTION:

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

Example 1:

Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL Example 2:

Input: 0->1->2->NULL, k = 4 Output: 2->0->1->NULL Explanation: rotate 1 steps to the right: 2->0->1->NULL rotate 2 steps to the right: 1->2->0->NULL rotate 3 steps to the right: 0->1->2->NULL rotate 4 steps to the right: 2->0->1->NULL

EXPLANATION:

看到题目后就知道应该是循环链表，因为最后大概率要首尾相连，干脆在一开始就首尾链接在一起然后再进行转换。
代码逻辑：
1.遍历链表，得到长度同时将首尾相连
2.需要移动的步数其实就是lenght-k-1，但是由于可能有length=1的情况发生，所以就不减1了。
3.开始移动需要的步数-1，这样就能拿到结果的前一个节点
4.将result的前一个进行解绑，并返回result。

SOLUTION:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        if(head==null) return null;
        ListNode root = head;
        int length = 1;
        while (head.next!=null){
            head = head.next;
            length++;
        }
        head.next = root;
        int step = length - k%length;
        while (step > 1){
            root = root.next;
            step--;
        }
        ListNode result = root.next;
        root.next = null;
        return result;
    }
}
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