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;
};

int main() {
    int node_len;
    while (cin >> node_len) {
        ListNode head{0, nullptr};
        ListNode* curr = &head;

        while (node_len--) {
            int value;
            cin >> value;
            auto* next = new ListNode{value, nullptr};
            curr->m_pNext = next;
            curr = next;
        }

        int k;
        cin >> k;

        auto slow = head.m_pNext;
        auto fast = head.m_pNext;
        while (k-- && fast) {
            fast = fast->m_pNext;
        }
        if (k <= 0) {
            while (fast) {
                fast = fast->m_pNext;
                slow = slow->m_pNext;
            } 
        }
        cout << slow->m_nKey << endl;
    }
}
// 64 位输出请用 printf("%lld")