61. Rotate List(Linked List-Medium)

1->2->3->4->5->NULL and k = 2

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* rotateRight(struct ListNode* head, int k) {
    int index = 0;
    int listlength = 1;
    struct ListNode *newlist = (struct ListNode *)malloc(sizeof(struct ListNode));
    struct ListNode *tail = (struct ListNode *)malloc(sizeof(struct ListNode));
    if(head == NULL){
        return NULL;
    }
    tail = head;
    newlist = head;
    while(tail->next){              //计算链表长度
        tail = tail->next;
        listlength++;
    }
    k = k % listlength;             //k可能大于链表的长度
    if(k == 0){                     //k等于链表长度,不旋转
        return head;
    }
    k = listlength - k;             //移动前面listlength-k个节点到右侧,后k个节点不动
    tail->next = head;              //尾节点连接首节点
    for(index; index < k-1; index++){      //找都新链表头
        newlist = newlist->next;
    }
    head = newlist->next;           //新链表头
    newlist->next = NULL;           //链表结尾赋值NULL
    return head;                    //返回链表头结点
}

/**
 * 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){
            return NULL;
        }
        ListNode *tail = head;
        int listlength = 1;
        while(tail->next){              //计算链表长度
            tail = tail->next;
            listlength++;
        }
        k = k % listlength;             //k可能大于链表的长度
        if(k == 0){                     //k等于链表长度,不旋转
            return head;
        }
        k = listlength - k;             //移动前面listlength-k个节点到右侧,后k个节点不动
        tail->next = head;              //尾节点连接首节点
        ListNode *newlist = head;
        for(int index = 0; index < k-1; index++){      //找都新链表头
            newlist = newlist->next;
        }
        head = newlist->next;           //新链表头
        newlist->next = NULL;           //链表结尾赋值NULL
        return head;                    //返回链表头结点
    }
};

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

class Solution(object):
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head:
            return None
        tail = head
        listlength = 1
        while tail.next:
            tail = tail.next
            listlength += 1
        k = k % listlength
        if k == 0:
            return head
        k = listlength - k
        tail.next = head
        newlist = head
        for each in range(k - 1):
            newlist = newlist.next
        head = newlist.next
        newlist.next = None
        return head