题解 | #输出单向链表中倒数第k个结点#

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){};
};

int main(){
    int n;
    while(cin >> n){
        ListNode *head = new ListNode(0);
        ListNode *cur = head;
        for(int i = 0; i < n; i++){
            int t;
            cin >> t;
            ListNode *temp = new ListNode(t);
            cur->m_pNext = temp;
            cur = cur->m_pNext;
        }
        head = head->m_pNext;
        
        int k;
        cin >> k;
        if(k > n || k == 0){
             cout << 0 << endl;
            continue;
        }
        ListNode *fast = head, *slow = head;
        for(int i = 0; i < k; i++){
            fast = fast->m_pNext;
        }
        while(fast != nullptr){
            fast = fast->m_pNext;
            slow = slow->m_pNext;
        }
        cout << slow->m_nKey << endl;;
    }
    return 0;
}