题解 | #找出单向链表中的一个节点到尾指针的距离为K#

#include <iostream>
using namespace std;
struct ListNode
{
int m_nKey;
ListNode* m_pNext;
};

ListNode *head,*r,*p,*d;
int pos;

int main() 
{
    cin >>pos;
    head = new ListNode;
    head->m_nKey = 1;
    head->m_pNext = NULL;
    r=head;
    //链表初始化
    for(int i=2 ; i<=7 ;i++)
    {
        p = new ListNode;
        p->m_nKey = i;
        p->m_pNext = NULL;
        r->m_pNext = p;
        r=p;
    }
    //输出链表，检查赋值是否正确
    // r=head;
    // while(r->m_pNext != NULL)
    // {
    //     cout << r->m_nKey <<" ";
    //     r=r->m_pNext;
    // }
    // cout << r->m_nKey;
    r=head; p = r->m_pNext; d = p->m_pNext;
    
    while(true)
    {
        p->m_pNext = r;
        if (d->m_pNext !=NULL)
        {r = p ; p = d; d = d->m_pNext;}
        else {d->m_pNext = p; break;}
    }
    head = d; //最后一个指针为头指针
    r=head;
    for(int i=1;i<=pos-1;i++)
    {
        r = r->m_pNext;
    }
    cout << r->m_nKey;


    
    return 0; 
}
// 64 位输出请用 printf("%lld")