我有一个链表的数组，我试图递归反转。当我调用函数反转时，它不会反转所有节点，而是反转几个节点。
反向功能似乎是删除第一个节点（基本情况）并用最后一个节点填充其位置（子情况结束）。我认为问题在于在 reverse_nodes 函数中调用for循环，但这似乎没有解决它。
这是一些输出..
pre-reverse function:
-----
group 0
alice, 2
-----
group 1
martin, 4
-----
group 2
keanu, 6
-----
group 3
miles, 8
post - reverse function
-----
group 0
miles, 8
-----
group 1
martin, 4
-----
group 2
keanu, 6
-----
group 3
miles, 8
我试图让它反转为：8,6,4,2
请注意我只包含相关的代码块，例如结构体系结构，头/尾构造，在读取二进制文件之前删除所有节点，将二进制文件读取到节点以及主要功能。我可以得到一些帮助，找出反向功能导致它不能完全反转的内容吗？谢谢你的时间。请参阅下面的代码！
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
typedef struct node
{
char name[20];
int size;
struct node *next;
}node;
node* head[4]={NULL,NULL,NULL,NULL};
node* tail[4]={NULL,NULL,NULL,NULL};
void wipe_nodes()
{
int i;
node *p=NULL;
for(i=4; i>0; i--)
{
p=head[i];
if(p == NULL)
{
printf("Nodes are clear!\n");
}
while(p != NULL)
{
delete_party(p->name, p->size); // cant call name and size
p = p -> next;
}
}
}
void bin_to_list(char *filename)
{
FILE *fp;
int ret;
fp = fopen(filename, "rb");
if (fp == NULL)
{
printf("Null file!\n");
return;
}
node temp;
//temp = (node *)malloc(sizeof(node));
while((ret = fread(&temp, sizeof(node), 1, fp) > 0))
{
printf("%s %d", temp.name, temp.size);
if(temp.size == 0)
{
printf("\nThat is not a valid command. Party not added!\n");
}
if(temp.size >= 1 && temp.size <= 2)
{
add_party(0, temp.name, temp.size);
}
else if(temp.size >= 3 && temp.size <= 4)
{
add_party(1, temp.name, temp.size);
}
else if(temp.size >= 5 && temp.size <= 6)
{
add_party(2, temp.name, temp.size);
}
else if(temp.size >= 7)
{
add_party(3, temp.name, temp.size);
}
}
fclose(fp);
return;
}
void reverse_nodes(node *p, node *q)
{
int i;
for(i=0; i<4; i++)
{
//node *p=head[i];
if(p == NULL)
{
printf("Error, no nodes!\n");
return;
}
if (p->next == NULL)
{
printf("LOL, only one node! Can't reverse!\n");
head[i] = p;
}
else
{
reverse_nodes(p->next, p);
}
p->next = q;
return;
int main(int argc, char *argv[])
{
int x, i;
read_to_list(argv[1]);
bin_to_list(argv[2]);
while (1)
{
fflush(stdin);
printf("\n\nEnter 1 to add a party\nEnter 2 to remove a party\nEnter 3 for the list of the party\nEnter 4 to change party size.\nEnter 5 to quit (write to .txt file).\nEnter 6 to read from bin file.\nEnter 7 to reverse the list.\n\n");
scanf("%d",&x);
char name[20];
int size;
switch(x)
{
case 1:
printf("\nParty Name: ");
scanf("%s", name);
printf("\nParty Size: ");
scanf("%d", &size);
if(size == 0)
{
printf("\nThat is not a valid command. Party not added!\n");
}
if(size >= 1 && size <= 2)
{
add_party(0, name, size);
}
else if(size >= 3 && size <= 4)
{
add_party(1, name, size);
}
else if(size >= 5 && size <= 6)
{
add_party(2, name, size);
}
else if(size >= 7)
{
add_party(3, name, size);
}
break;
case 2:
printf("\nSize of party to delete: ");
scanf("%i", &size);
delete_party(NULL, size);
break;
case 3:
list_parties();
break;
case 4:
change_partysize(name, size);
break;
case 5:
write_to_file(argv[1]);
write_to_bin(argv[2]);
exit(0);
break;
case 6:
wipe_nodes();
bin_to_list(argv[2]);
break;
case 7:
for(i=0; i<4; i++)
{
reverse_nodes(head[i], NULL);
}
break;
default:
continue;
}
}
}
答案 0 :(得分：0)
#include<stdio.h>
#include<stdlib.h>
/* Link list node */
struct Node
{
int data;
struct Node* next;
};
/* Function to reverse the linked list */
static void reverse(struct Node** head_ref)
{
struct Node* prev = NULL;
struct Node* current = *head_ref;
struct Node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
/* Function to push a node */
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print linked list */
void printList(struct Node *head)
{
struct Node *temp = head;
while(temp != NULL)
{
printf("%d ", temp->data);
temp = temp->next;
}
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 85);
printf("Given linked list\n");
printList(head);
reverse(&head);
printf("\nReversed Linked list \n");
printList(head);
getchar();
}
答案 1 :(得分：0)
您的reverse_nodes函数只是设置next的{{1}}成员，而不是当前节点，因此当它到达列表末尾时，它将设置node <} p}。链接列表中前一个成员的最后一个成员，但保持最后一个成员不变。
这对我有用：
next编辑：
递归通常是一个坏主意，因为你可能最终会破坏堆栈 - 如果可能的话，最好尝试在单个（或多个，如果需要的话）函数中做你需要做的事情，在这种情况下很容易。