题目英文

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

题目中文

给定一个链表，旋转链表，将链表每个节点向右移动 k 个位置，其中 k 是非负数。

示例 1:

输入: 1->2->3->4->5->NULL, k = 2

输出: 4->5->1->2->3->NULL

解释:

向右旋转 1 步: 5->1->2->3->4->NULL

向右旋转 2 步: 4->5->1->2->3->NULL

示例 2:

输入: 0->1->2->NULL, k = 4

输出: 2->0->1->NULL

解释:

向右旋转 1 步: 2->0->1->NULL

向右旋转 2 步: 1->2->0->NULL

向右旋转 3 步: 0->1->2->NULL

向右旋转 4 步: 2->0->1->NULL

算法实现

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
 
public class Solution
{
    public ListNode RotateRight(ListNode head, int k)
    {
        if (head == null || k == 0)
            return head;

        int len = GetLength(head);
        int index = len - k%len;

        if (index == len)
            return head;

        ListNode temp1 = head;
        ListNode temp2 = head;
        for (int i = 0; i < index - 1; i++)
        {
            temp1 = temp1.next;
        }
        head = temp1.next;
        temp1.next = null;

        temp1 = head;
        while (temp1.next != null)
        {
            temp1 = temp1.next;
        }
        temp1.next = temp2;
        return head;
    }

    public int GetLength(ListNode head)
    {
        ListNode temp = head;
        int i = 0;
        while (temp != null)
        {
            i++;
            temp = temp.next;
        }
        return i;
    }
}