题解 | #输出单向链表中倒数第k个结点#

#include <stdio.h>
#include <stdlib.h>
typedef struct Node {
    int key;
    struct Node* next;
}Node;

int main() {
    int n;
    int key;
    while(scanf("%d", &n) != EOF) {
        Node* head = (Node*)malloc(sizeof(Node));
        head->next=NULL; //头节点
        for(int i=0; i<n; i++) {
        scanf("%d", &key);
        Node* newnode = (Node*)malloc(sizeof(Node));
		  //头插法就不需要快慢指针了
        newnode->key=key;
        newnode->next=head->next;
        head->next=newnode;
    }
    
    int k;
    scanf("%d",&k);
    Node*p=head;
    for(int i=0; i<k; i++) {
        head=head->next;
    }
    printf("%d\n",head->key);
    }//节点个数
    
}