給定一個連結串列，旋轉連結串列，將連結串列每個節點向右移動 k 個位置，其中 k 是非負數。

示例 1:

輸入: 1->2->3->4->5->NULL, k = 2
輸出: 4->5->1->2->3->NULL
解釋:
向右旋轉 1 步: 5->1->2->3->4->NULL
向右旋轉 2 步: 4->5->1->2->3->NULL

示例 2:

輸入: 0->1->2->NULL, k = 4
輸出:
 
 2->0->1->NULL
解釋:
向右旋轉 1 步: 2->0->1->NULL
向右旋轉 2 步: 1->2->0->NULL
向右旋轉 3 步: 0->1->2->NULL
向右旋轉 4 步: 2->0->1->NULL

C

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* rotateRight(struct ListNode* head, int k) 
{
    if(NULL==head || NULL==head->next)
    {
        return head;
    }
    struct ListNode* last=head;
    int count=1;
    while(last->next)
    {
        count++;
        last=last->next;
    }
    k%=count;
    if(0==k)
    {
        return head;
    }
    int i=1;
    struct ListNode* tmp=head;
    while(i<count-k)
    {
        tmp=tmp->next;
        i++;
    }
    struct ListNode* start=tmp->next;
    tmp->next=NULL;
    last->next=head;
    return start;    
}
 

C++

/**
 * 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 || NULL==head->next)
        {
            return head;
        }
        int count=1;
        ListNode* last=head;
        while(last->next)
        {
            count++;
            last=last->next;
        }
        k%=count;
        if(0==k)
        {
            return head;
        }
        ListNode* tmp=head;
        int i=1;
        while(i<count-k)
        {
            tmp=tmp->next;
            i++;
        }
        ListNode* start=tmp->next;
        tmp->next=NULL;
        last->next=head;
        return start;
    }
};
 

python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if None==head or None==head.next:
            return head
        count=1
        last=head
        while last.next:
            last=last.next
            count+=1
        k%=count
        if 0==k:
            return head
        tmp=head
        i=1
        while i<count-k:
            tmp=tmp.next
            i+=1
        start=tmp.next
        tmp.next=None
        last.next=head
        return start