Rotate List

Description:

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->nullptr and k = 2, return 4->5->1->2->3->nullptr.
分析
先遍历一遍，得出链表长度 len，注意 k 可能大于 len，因此令 k% = len。将尾节点 next 指针
指向首节点，形成一个环，接着往后跑 len − k 步，从这里断开，就是要求的结果了。

#include <iostream>#include <limits.h>using namespace std;class LNode{public:    int val;    LNode *next;    LNode(int x):val(x),next(nullptr) { };};class Solution{public:    LNode *rotateList(LNode *L,int k)    {        if ( L == nullptr || k == 0)            return L;        int len = 1;        LNode *p = L;   //p used to traverse the list        //cout<<p->val;        while(p->next)        {            ++len;            p = p->next;        }        //cout<<len;        k = len - k%len;        //cout<<k;        p->next = L;    // end to end        while (k)        {            p = p->next;            --k;        }        L = p->next;    // new head node        p->next = nullptr;  //break the ring        return L;    }};int main(void){    //1->2->3->4->5->nullptr;    LNode *List = new LNode(INT_MIN);   //create head node    LNode *ptr = List;    int n;  //input the number    cout<<"input the number(non-negtive): ";    cin>>n;    //create the list    cout<<"input the value"<<endl;    for (int i = 0; i < n; ++i)    {        int value;        cin>>value;        ptr->next = new LNode(value);        ptr = ptr->next;        cout<<ptr->val;        if (i < n-1)            cout<<"->";    }    cout<<endl;    Solution solution;    int k = 2;    LNode *ret = solution.rotateList(List->next,k);    LNode *pointer = ret;    //output and delete the list    delete List;    //delete head node    while(pointer)    {        LNode *del = pointer;        cout<<pointer->val;        if (pointer->next != nullptr)            cout<<"->";        pointer = pointer->next;        delete del;    }    cout<<endl;    return 0;}