leetcode rotate-list_leetcode rotate-list c++-优快云博客

本文链接：https://blog.youkuaiyun.com/pynash123/article/details/84572262

本文介绍了一种链表右旋算法的实现方法，通过双指针技巧完成链表的旋转，确保了即使当旋转次数超过链表长度时也能正确处理。文章提供了完整的C++代码示例，并展示了如何构造链表及输出旋转后的链表。

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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->NULLand k =2,
return4->5->1->2->3->NULL.

/**
 * 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(NULL == head || k == 0)
            return head;
        ListNode *Head = new ListNode (-1);
        ListNode * cur = Head, *fast = Head, *tmp = head;
        int length = 0;
        
        Head->next = head;
        while(NULL != tmp)
        {
            tmp = tmp->next;
            length ++;
        }
        k = k%length;//k比链表长度还大
        if(k == 0)
            return head;
        for (int i=0;i<k;i++)
            fast = fast->next;
        while(fast->next != NULL)
        {
            cur = cur->next;
            fast = fast->next;
        }
        fast->next = Head->next;
        Head->next = cur->next;
        cur->next = NULL;
        return Head->next;
    }
};

测试

#include"head.h"
class Solution {
public:
    ListNode *rotateRight(ListNode *head, int k) 
    {
        if(NULL == head || k == 0)
            return head;
        ListNode *Head = new ListNode (-1);
        ListNode * cur = Head, *fast = Head, *tmp = head;
        int length = 0;
        
        Head->next = head;
        while(NULL != tmp)
        {
            tmp = tmp->next;
            length ++;
        }
        k = k%length;
        if(k == 0)
            return head;
        for (int i=0;i<k;i++)
            fast = fast->next;
        while(fast->next != NULL)
        {
            cur = cur->next;
            fast = fast->next;
        }
        fast->next = Head->next;
        Head->next = cur->next;
        cur->next = NULL;
        return Head->next;
    }
};

int main()
{
    int A[] = {1,2,3,4,5};
    vector<int> vec(A, A+5);
    ListNode *head = makelist(vec);
    printlist(head);
    Solution s;
    ListNode * p = s.rotateRight(head, 5);
    printlist(p);
}