LeetCode刷题笔记（链表）：rotate-list

- 日理万妓 2022-06-01 01:58 152阅读 0赞

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*  **转载请注明作者和出处：[http://blog.csdn.net/u011475210][http_blog.csdn.net_u011475210]**
 *  **代码地址：[https://github.com/WordZzzz/Note/tree/master/LeetCode][https_github.com_WordZzzz_Note_tree_master_LeetCode]**
 *  **刷题平台：[https://www.nowcoder.com/ta/leetcode][https_www.nowcoder.com_ta_leetcode]**
 *  **题  库：Leetcode经典编程题**
 *  **编  者：WordZzzz**

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*   *  题目描述
     *  解题思路
     *  C版代码实现

## 题目描述 ##

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

For example:  
Given1->2->3->4->5->NULLand k =2,  
return4->5->1->2->3->NULL.

## 解题思路 ##

*  遍历链表得到链表长度并定位fast到尾结点；
 *  根据链表长度，slow找到新的尾结点，slow的下一个结点便是新的头结点因为k可能比len大，所以需要取余数；
 *  原先的尾结点与原先的头节点相连，dummy的next指针更新指向新的头结点，同时断开新的尾节点与新的头结点之间的联系。

## C++版代码实现 ##

/**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode *rotateRight(ListNode *head, int k) {
            if(k == 0 || head == NULL || head->next == NULL)
                return head;
    
            ListNode *dummy = new ListNode(0);
            dummy->next = head;
            ListNode *fast = head;
            ListNode *slow = head;
            int len;
            //计算链表长度，同时fast指向尾结点
            for(len = 1; fast->next != NULL; ++len)
                fast = fast->next;
            //根据得到的链表长度，找到需要切开的结点
            for(int i = 1; i < len - k % len; ++i)
                slow = slow->next;
            fast->next = dummy->next;   //尾结点连接到头节点，形成环状链表
            dummy->next = slow->next;   //fast的下一个结点作为新的头结点
            slow->next = NULL;          //slow结点和新的头结点断开
    
            return dummy->next;
        }
    };

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[http_blog.csdn.net_u011475210]: http://blog.csdn.net/u011475210
[https_github.com_WordZzzz_Note_tree_master_LeetCode]: https://github.com/WordZzzz/Note/tree/master/LeetCode
[https_www.nowcoder.com_ta_leetcode]: https://www.nowcoder.com/ta/leetcode