题解 | #输出单向链表中倒数第k个结点#

#include <iostream>
using namespace std;

struct ListNode { //链表结点
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(NULL) {} //初始化
};

ListNode* FindKthToTail(ListNode* pHead, int k) {
    ListNode* fast = pHead, *slow = pHead;
    for (int i = 0; i < k; i++) {
        if (fast == nullptr) {
            return slow = nullptr;
        }
        fast = fast->next;
    }
    while (fast != nullptr) {
        fast = fast->next;
        slow = slow->next;
    }
    return slow;
}

int main() {
    int n;
    while (cin >> n) { //输入n
        int val;
        cin >> val;
        ListNode* head = new ListNode(val); //链表第一个结点
        ListNode* p = head;
        for (int i = 1; i < n; i++) { //输入链表后续结点
            cin >> val;
            ListNode* q = new ListNode(val);
            p->next = q; //连接
            p = p->next;
        }
        int k;
        cin >> k; //输入k

        if (k == 0) //k等于0直接输出0
            cout << 0 << endl;
        else {
            p = FindKthToTail(head, k); //找到第k个结点
            if (p != nullptr) //返回不为null才能输出
                cout << p->val << endl;
        }
    }
}
// 64 位输出请用 printf("%lld")