题解 | #输出单向链表中倒数第k个结点#

#include <iostream>
using namespace std;

typedef struct ListNode{
    int m_nKey;
    ListNode* m_pNext;
    ListNode(int x) : m_nKey(x), m_pNext(nullptr) {};
} *List;

int main() {
    int num;
    while (cin>>num) {
    int a;
    cin>>a;
    List head = new ListNode(a);
    List p = head;
    for(int i = 1;i<num;i++){
        cin>>a;
        List t = new ListNode(a);
        p->m_pNext = t;
        p = t;
    }
    p = head;
    List pp = head;
    int k;
    cin>>k;
    while(k>0&&pp){
        pp = pp->m_pNext;
        k--;
    }
    //if(!pp) return 0;
    while(pp){
        pp = pp->m_pNext;
        p=p->m_pNext;
    }
    cout<<p->m_nKey<<endl;
    }
    

}
// 64 位输出请用 printf("%lld")

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