leetcode_61_Rotate List

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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.

注意一点，K可能非常大，比链表的长度大得多，但是因为是循环右移，所以实际上只要循环右移K%Length位（Length为链表长度）。

//没理解好题意，写成了将右边k个数字移到前面,但是也能通过//注意一点，K可能非常大，比链表的长度大得多，但是因为是循环右移，所以实际上只要循环右移K%Length位（Length为链表长度）。class Solution {public:    ListNode *rotateRight(ListNode *head, int k) {        if(head==NULL)            return head;int len = getlistlen(head);k = k%len;int n = len - k;if(k==0)return head;ListNode *first=head;for(int i=0; i<n-1; i++){first = first->next;}ListNode *head2 = first->next;first->next = NULL;ListNode *second = head2;for(int i=0; i<k-1; i++){second = second->next;}second->next = head;return head2;    }private:int getlistlen( ListNode *head){int len=0;while(head){len++;head = head->next;}return len;}};

//另一种写法class Solution {public:    ListNode *rotateRight(ListNode *head, int k) {        if (head == NULL || head->next == NULL || k == 0) {            return head;        }                int length = 0;        ListNode *ptr = head, *tail = head;        while (ptr != NULL) {            length++;            tail = ptr;            ptr = ptr->next;         }        k %= length;                ptr = head;        for (int i = 0; i < length - k - 1; i++) {            ptr = ptr-> next;        }                tail->next = head;        head = ptr->next;        ptr->next = NULL;                return head;    }};

#include<iostream>using namespace std;#define N 5struct ListNode {int val;ListNode *next;ListNode(int x) : val(x), next(NULL) {}};//没理解好题意，写成了将右边k个数字移到前面,但是也能通过class Solution {public:    ListNode *rotateRight(ListNode *head, int k) {        if(head==NULL)            return head;int len = getlistlen(head);k = k%len;int n = len - k;if(k==0)return head;ListNode *first=head;for(int i=0; i<n-1; i++){first = first->next;}ListNode *head2 = first->next;first->next = NULL;ListNode *second = head2;for(int i=0; i<k-1; i++){second = second->next;}second->next = head;return head2;    }private:int getlistlen( ListNode *head){int len=0;while(head){len++;head = head->next;}return len;}};ListNode *creatlist(){ListNode *head = NULL;ListNode *p;for(int i=0; i<N; i++){int a;cin>>a;p = (ListNode*) malloc(sizeof(ListNode));p->val = a;p->next = head;head = p;}return head;}int main(){ListNode *list = creatlist();Solution lin;int a;cin>>a;ListNode *outlist = lin.rotateRight ( list,a );for(int i=0; i<N; i++){cout<<outlist->val;outlist = outlist->next;}}

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