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;
    while (cin >> n) {
        int num, k;
        ListNode* T = (ListNode*)malloc(sizeof(ListNode));
        T->m_pNext = nullptr;
        ListNode* p = T;
        for (int i = 0; i < n; i++) {
            cin >> num;
            p->m_pNext = (ListNode*)malloc(sizeof(ListNode));
            p = p->m_pNext;
            p->m_nKey = num;
            p->m_pNext = nullptr;
        }
        cin >> k;
        p = T->m_pNext;
        ListNode* q = T->m_pNext;
        while (k) {
            q = q->m_pNext;
            k--;
        }
        while (q) {
            p = p->m_pNext;
            q = q->m_pNext;
        }
        cout << p->m_nKey << endl;
    }

}
// 64 位输出请用 printf("%lld")