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(int x):val(x),next(nullptr){}
};
ListNode* findKthTOtail(ListNode* head, int k){
    ListNode *fast = head, *slow = head;
    while(k--){
        if(fast != nullptr) fast = fast->next;
        else return nullptr;
    }
    while(fast != nullptr){
        fast = fast->next;
        slow = slow->next;
    }
    return slow;
}
int main() {
    int n;
    while(cin>>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;
            p->next = new ListNode(val);
            p = p->next;
        }
        int k;
        cin>>k;
        if(k == 0) cout<< 0<< endl;
        else {
            p = findKthTOtail(head, k);
            if(p != nullptr) cout<<p->val<<endl;
        }
    }
    return 0;
}