题解 | #输出单向链表中倒数第k个结点#

#include <cstddef>
#include <iostream>
using namespace std;
struct ListNode
{
    int m_nKey;
    ListNode* m_pNext;
    ListNode(int x): m_nKey(x),m_pNext(NULL){}  //初始化
};

ListNode* insertNode(int n, ListNode *head,int k){
    ListNode *p=head;
    if(n<k)return NULL;
   for(int i=0;i<n-k;i++){
    p=p->m_pNext;
   }

    return p;
}
int main() {
    int n;
    while(cin>>n)
    {int data;
    cin>>data;
    ListNode *head=new ListNode(data);
    ListNode *p=head;
    head->m_pNext=NULL;
    for(int i=1;i<n;i++){
        cin>>data;
        ListNode *q=new ListNode(data);
        p->m_pNext=q;
        p=p->m_pNext;
    }
    int num;
    cin>>num;
   
    if(num==0)cout<<0<<endl;
    else {
       p =insertNode(n,head, num);
       if(p!=NULL)cout<<p->m_nKey<<endl;
    }}

}
// 64 位输出请用 printf("%lld")