题解 | #输出单向链表中倒数第k个结点#

#include <iostream>

using namespace std;

struct ListNode
{
    int m_nKey;
    ListNode* m_pNext;
    //构造函数
    ListNode():m_nKey(0),m_pNext(nullptr){};
    ListNode(int x):m_nKey(x),m_pNext(nullptr){};
};

int main()
{
    int num,node;
    while(cin>>num)
    {
        //链表的头
        ListNode* head=new ListNode();
        ListNode* pre_head = head;
        //初始化链表
        while(num--)
        {
            cin>>node;
            ListNode* next=new ListNode(node);
            head->m_pNext = next;
            head = head->m_pNext;
        }
        //输入需要查询的位置:num
        cin>>num;
        ListNode* first = pre_head;//指向了输入的头；
        if(num<=0)
        {
            cout<<"0"<<endl;
            continue;
        }
        for(int i=0; i<num; i++)
        {
            if(first->m_pNext != nullptr)
            {
                first = first->m_pNext;
            }
            else 
            {
                cout<<"0"<<endl;
                continue;
            }
        }
        ListNode* result = pre_head->m_pNext;
        while(first->m_pNext != nullptr)
        {
            result = result->m_pNext;
            first = first->m_pNext;
        }
        cout<<result->m_nKey<<endl;
    }

    return 0;
}