旋转链表

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给定一个链表，旋转链表，使得每个节点向右移动k个位置，其中k是一个非负数

样例

给出链表1->2->3->4->5->null和k=2

返回4->5->1->2->3->null

#include <iostream>using namespace std;//Definition for singly-linked list.struct ListNode {    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};class Solution {public:/*** @param head: the list* @param k: rotate to the right k places* @return: the list after rotation*/ListNode* Init(ListNode* head){int val = 0;ListNode* p = NULL;while (true){cin >> val;if (-1 == val){return head;}if (NULL == head){head = new ListNode(val);p = head;}else{p->next = new ListNode(val);p = p->next;}}}int getListLength(ListNode* head){int count = 0;if (NULL == head){return 0;}else{ListNode* p = head;while (p!=NULL){count++;p = p->next;}}return count;}ListNode *rotateRight(ListNode *head, int k) {// write your code hereint len = getListLength(head);if (0 == len || NULL == head){return 0;}k = k%len;if (k == len || 0 == k)//反转的次数和长度一样或者反转的长度为0，直接返回{return head;}ListNode* tail = NULL;ListNode* prev = head;ListNode* p = head;int count = len - k;count--;//因为这个链表没有头结点，头就是第一个节点，所以要先减一while (count--){p = p->next;}tail = p->next;p->next = NULL;p = tail;while (p->next!=NULL){p = p->next;}p->next = prev;return tail;}void Out(ListNode* head){if (head !=NULL){ListNode* p = head;while (p){cout << p->val << " ";p = p->next;}}cout << endl;}};int main(){ListNode* head = NULL;Solution s;head = s.Init(head);s.Out(head);int k = 0;cin >> k;head = s.rotateRight(head, k);s.Out(head);return 0;}

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