题解 | #输出单向链表中倒数第k个结点#

#include<iostream>
using namespace std;
struct Node
{
    int iData;
    struct Node* pNext;
};
int n, iData, k;
int main()
{
    while(cin>>n)
    {       
        struct Node* pHead = nullptr;//头指针
        struct Node* pEnd= nullptr;  //尾指针
        
        for(int i=0; i<n; i++)
        {
            cin>>iData;
            struct Node* pTemp = new struct Node; //生成当前节点
            if(pTemp!=nullptr)  
            {
                pTemp->iData = iData;  //节点赋值
                pTemp->pNext = nullptr;
                if(pHead== nullptr)  //连上去
                {
                    pHead = pTemp;
                    pEnd = pTemp;
                }    
                else
                {
                    pEnd->pNext = pTemp;
                    pEnd = pTemp;
                }
            }  
        }
        cin>>k;
        int i = 0; 
        k = n-k;   //倒数第k个节点就是正数下标为（n-k）的节点
        struct Node* pt = pHead;  //遍历
        while(i < k)
        {
            pt = pt->pNext;
            i++;
        }
        cout<<pt->iData<< endl;
    }
    return 0;
}