用链表依次存储1,2,3,4并实现反序输出

可以使用单向链表来依次存储1,2,3,4，具体实现如下：

1. 定义链表节点结构体

typedef struct ListNode {
    int val; // 节点值
    struct ListNode *next; // 指向下一个节点的指针
} ListNode;

2. 创建链表并添加节点

ListNode* createList() {
    ListNode *head = NULL, *tail = NULL;
    for (int i = 1; i <= 4; i++) {
        ListNode *node = (ListNode*)malloc(sizeof(ListNode));
        node->val = i;
        node->next = NULL;
        if (head == NULL) {
            head = tail = node;
        } else {
            tail->next = node;
            tail = node;
        }
    }
    return head;
}

3. 反序输出链表

void reversePrint(ListNode *head) {
    if (head == NULL) {
        return;
    }
    reversePrint(head->next);
    printf("%d ", head->val);
}

完整代码如下：

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

typedef struct ListNode {
    int val;
    struct ListNode *next;
} ListNode;

ListNode* createList() {
    ListNode *head = NULL, *tail = NULL;
    for (int i = 1; i <= 4; i++) {
        ListNode *node = (ListNode*)malloc(sizeof(ListNode));
        node->val = i;
        node->next = NULL;
        if (head == NULL) {
            head = tail = node;
        } else {
            tail->next = node;
            tail = node;
        }
    }
    return head;
}

void reversePrint(ListNode *head) {
    if (head == NULL) {
        return;
    }
    reversePrint(head->next);
    printf("%d ", head->val);
}

int main() {
    ListNode *head = createList();
    reversePrint(head);
    return ;
}