用c原因完成一段代码要求有输入和输出利用单链表原来的节点空间逆转一个单链表

#include<stdio.h>

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

void reverseList(struct Node** head_ref) { struct Node* prev = NULL; struct Node* current = head_ref; struct Node next = NULL; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } *head_ref = prev; }

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

int main() { struct Node* head = NULL; struct Node* second = NULL; struct Node* third = NULL;

head = (struct Node*)malloc(sizeof(struct Node)); 
second = (struct Node*)malloc(sizeof(struct Node)); 
third = (struct Node*)malloc(sizeof(struct Node)); 

head->data = 1; 
head->next = second; 

second->data = 2; 
second->next = third; 

third->data = 3; 
third->next = NULL; 

printf("Original List:\n"); 
printList(head); 

reverseList(&head); 

printf("\nReversed List:\n"); 
printList(head); 

return 0; 

}

/* 输入： 无 输出： Original List: 1 2 3 Reversed List: 3 2 1 *