链表-旋转链表-中等

描述
给定一个链表，旋转链表，使得每个节点向右移动k个位置，其中k是一个非负数
您在真实的面试中是否遇到过这个题？  
样例
给出链表1->2->3->4->5->null和k=2

返回4->5->1->2->3->null

题目链接

程序

/**
 * Definition of singly-linked-list:
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *        this->val = val;
 *        this->next = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param head: the List
     * @param k: rotate to the right k places
     * @return: the list after rotation
     */
    ListNode * rotateRight(ListNode * head, int k) {
        // write your code here
        //版本一
        /*
        if(head == NULL || k == 0 || head->next == NULL)
            return head;
        ListNode *cur = head;
        int size = 0;
        while(cur){//计算链表得到全部长度
            size++;
            cur = cur->next;
        }
        //cout << "总长度 " << size << endl;
        if(k % size != 0){
            cur = head;
            int curList = size - k % size;//断开处的位置
            //cout << "断开处 " << curList << endl;
            for(int i = 1; i < curList; i++){//找到断开处
                cur = cur->next;
            }
            ListNode *afterList = cur->next;//后半段链表的头结点
            //cout << "后半段链表的头结点 " << afterList->val << endl;
            cur->next = NULL;//前半段链表末尾指向NULL
            //cout << "前半段链表的尾结点 " << cur->val << endl;
            cur = afterList;
            while(cur->next){//找到后半段的最后一个节点
                cur = cur->next;
            }
            cur->next = head;//让最后一个节点指向前半段的头结点
            return afterList;
        }
        else{
            return head;
        }
        */
        
        //版本二
        if(head == NULL)
            return head;
        int len = 0;
        for(ListNode *node = head; node != NULL; node = node->next)
            len++;
        k = k%len;
        ListNode *fast = head;
        for(int i = 0; i < k; i++){
            fast = fast->next;
        }
        ListNode *slow = head;
        while(fast->next != NULL){
            slow = slow->next;
            fast = fast->next;
        }
        fast->next = head;
        head = slow->next;
        slow->next = NULL;
        return head;
    }
};