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(): m_nKey(0), m_pNext(nullptr) {};
    ListNode(int x): m_nKey(x), m_pNext(nullptr) {};
};

ListNode* getNode(ListNode* node, int& node_num) {
    if (node == NULL) return NULL;
    ListNode* head = getNode(node->m_pNext, node_num);
    if (--node_num == 0) return node;
    else return head;
}

int main() {
    int node_num, node;
    while (cin >> node_num) {
        ListNode* head = new ListNode(); //正序构建链表
        ListNode* pre_head = head;
        while (node_num--) {
            cin >> node;
            ListNode* next = new ListNode(node);
            head->m_pNext = next;
            head = next;
        }
        cin >> node_num;
        ListNode* rec = getNode(pre_head->m_pNext,
                                node_num); //忘记链表长度，递归找到指定节点
        if (rec != NULL) cout << rec->m_nKey << endl;
        else cout << "0" << endl;
    }
    return 0;
}