Python怎么将链表旋转到右侧

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For example:

Given

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

return

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

题目： 给定一个链表，通过k(非负)节点将链表旋转到右侧。

解读： 一直感觉这个题目的表述有些问题。我是这样理解才做对题的：保留后k个节点，比如example给出的k=2，那么保留4->5，前面的3个节点1->2->3右移到后面，形成新链表4->5->1->2->3。除此之外，k值也有可能是大于链表长度的，需要取余，获得真实的位移。

Language : c

/** * 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NULLreturn head;                    //返回链表头结点}

Language : cpp

/** * 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NULLreturn head;                    //返回链表头结点}};

Language:python

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):def rotateRight(self, head, k):"""        :type head: ListNode        :type k: int        :rtype: ListNode        """if not head:return Nonetail = head        listlength = 1while tail.next:            tail = tail.next            listlength += 1k = k % listlengthif k == 0:return head        k = listlength - k        tail.next = head        newlist = headfor each in range(k - 1):            newlist = newlist.next        head = newlist.next        newlist.next = Nonereturn head

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