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;
};
ListNode* move(ListNode* p, int k) {
    for (int a = 0; a < k; a++) {
        p = p->m_pNext;
    }
    return p;
}
void pro(int a) {
    ListNode* head, *curr, *last;
    head = new ListNode();
    curr = last = head;
    for (int b = 0; b < a; b++) {
        cin >> last->m_nKey;
        last->m_pNext = new ListNode();
        last = last->m_pNext;
    }
    cin >> a; //a为k
    while (move(curr, a) != last) {
        curr = curr->m_pNext;
    }
    cout << curr->m_nKey << endl;
}
int main() {
    int a;
    while (cin >> a) { //a为链表长度n
        pro(a);
    }
}