题解 | #输出单向链表中倒数第k个结点#

#include <stdio.h>
#include <stdlib.h>

struct ListNode
{
    int m_nKey;
   struct ListNode *m_pNext;
};

struct ListNode* create_node(void)
{
    struct ListNode *node = (struct ListNode *)malloc(sizeof(struct ListNode *));
    node->m_pNext=NULL;
    node->m_nKey=0;

    return node;
}

int main() {
    int a, b,i,j;
    

    while (scanf("%d", &a) != EOF) { 
        struct ListNode* newNode=NULL,*end=NULL,*head=NULL;
        for(i=0;i<a;i++)
        {
            // if(i==a)
            // {
            //     scanf("%d",&b);
            //     break;
            // }
            struct ListNode* newNode=create_node();
            scanf("%d",&newNode->m_nKey);
            if(head==NULL)
                head=newNode;
            else 
                end->m_pNext=newNode;
            end=newNode;
            // printf("%d",newNode->m_nKey);
        }
        if(end!=NULL)
            end->m_pNext=NULL;

        scanf("%d",&b);
        j=a-b;
        while(j)
        {
            head=head->m_pNext;
            j--;

        }
        printf("%d\n",head->m_nKey);
        
    }
    return 0;
}

注意单链表的创建