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;
};

int main() {
    int n,num,k;
    while (cin>>n) {
        ListNode* head = nullptr;
        ListNode* tail = nullptr;

        for (int i = 0; i < n; i++) {
            cin>>num;
            ListNode* node = new ListNode;
            node->m_nKey = num;
            node->m_pNext = nullptr;
            if (head == nullptr){
                head = node;
            } else{
                tail->m_pNext = node;
            }
            tail = node;
        }
        cin>>k;
        ListNode* i = head;
        ListNode* j = head;
        k = k - 1;
        while(k--){
            if (j == nullptr) break;
            j = j->m_pNext;
        }
        if (j != nullptr){
            while(j != tail){
                i = i->m_pNext;
                j = j->m_pNext;
            }
            cout<<i->m_nKey<<endl;
        }
    }
}