Leetcode Rotate List

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

.

创建一个哨兵节点phead，pthread->next=head,第一次遍历求出链表的长度len以及链表的最后节点tail，k=k%len,通过第二次遍历找到前半段的最后一个节点curr。如果curr==tail，那么说明k%len为0，只需返回head；否则phead->next = tail,tail->next=curr->next;curr->next = NULL(不加的还结果为循环列表，导致leetcode判定会超时);返回pthread->next.

代码如下：

/**
 * 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(head == NULL || head->next == NULL ||  k == 0)
            return head;
        ListNode* phead = new ListNode(0);
        ListNode* curr= head,*tail = head;
        
        int len =0;
        while(tail->next != NULL)
        {
            tail = tail->next;
            len++;
        }
        len++;
        
        k = k%len;
        if(k == 0 )
            return head;
            
        int count = 0;
        while( count < len-k-1)
        {
            curr = curr->next;
            count++;
        }
        

        phead->next = curr->next;
        curr->next = NULL;
        tail->next = head;
        return phead->next;
    }
};