题解 | 输出单向链表中倒数第k个结点

#include <iostream>
using namespace std;

struct ListNode{
     int val;
     ListNode* next;
     ListNode():val(-1),next(nullptr){}
     ListNode(int x):val(x),next(nullptr){}
     ListNode(int x,ListNode* next):val(x),next(next){}
};

int getnode(ListNode* head,int k){
    ListNode* fast = head;
    ListNode* slow = head;
    for(int i=0;i<k;i++){
        fast = fast->next;
    }
    while(fast){
        fast = fast->next;
        slow = slow->next;
    }
    return slow->val;
}

int main() {
    int node_num;
    int num;
    int ret;
    while(cin >> node_num){
    ListNode* head = new ListNode();
    ListNode* cur = nullptr;
    ListNode* ptr = head;
    for(int i=0;i<node_num;i++){
        //ListNode* ptr = head;
        cin >> num;
        cur = new ListNode(num);
        ptr->next = cur;
        ptr = cur;
    }
    cur = head->next;
    cin >> num;

    ret = getnode(cur,num);
    cout << ret << endl;
    }
    return 0;
}

利用快慢指针进行求解