对于一个不带头结点的单链表L，设计一个递归算法逆置所有结点。编写完整的程序并进行测试，分析其时间复杂度。使用c99编写该脚本

我可以回答这个问题。以下是逆置单链表的递归算法的C99代码：

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

struct Node {
    int data;
    struct Node* next;
};

void reverseList(struct Node* curr, struct Node* prev, struct Node** head) {
    if (curr->next == NULL) {
        *head = curr;
        curr->next = prev;
        return;
    }
    struct Node* next = curr->next;
    curr->next = prev;
    reverseList(next, curr, head);
}

void insert(struct Node** head, int data) {
    struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = data;
    newNode->next = *head;
    *head = newNode;
}

void printList(struct Node* head) {
    while (head != NULL) {
        printf("%d ", head->data);
        head = head->next;
    }
    printf("\n");
}

int main() {
    struct Node* head = NULL;
    insert(&head, 5);
    insert(&head, 4);
    insert(&head, 3);
    insert(&head, 2);
    insert(&head, 1);
    printf("Original list: ");
    printList(head);
    reverseList(head, NULL, &head);
    printf("Reversed list: ");
    printList(head);
    return 0;
}

该算法的时间复杂度为O(n)，其中n是链表中的节点数。