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LeetCode || Rotate List

Rotate List

 题目链接

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->NULL

 and k = 

2

,

return 

4->5->1->2->3->NULL

.

<span style="font-size:14px;">/**
 * 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)//k取值<=0 , 链表中没有节点以及只有一个节点的情况。
            return head;
        
        //计算链表的长度,为了除了k>链表长度时候的情况。
        ListNode* pfront = head;
        int count = 0;
        while(pfront != NULL)
        {
            count++;
            pfront= pfront->next;
        }
        k = k % count;//处理k的值
        
        //将链表向后移动,移动k位。
        count = 0;
        pfront = head;
        while(pfront != NULL && count < k)
        {
            pfront = pfront->next;
            count++;
        }
        //如果没有移动,则表示链表不需要reverse
        if(pfront == head)
            return head;
            
        //设置节点p,用来记录需要从哪里开始截断链表
        ListNode* p = head;
        while(p->next!=NULL && pfront->next !=NULL)
        {
            p = p->next;
            pfront = pfront->next;
        }

        //更改新的头指针
        ListNode* newHead = p->next;
        pfront->next = head;
        p->next = NULL;
        
        return newHead;
    }
};</span>