我正在尝试使用C实现Hierholzer的算法。我为使用双链表实现的简单堆栈制作了push函数,但是即使起始节点为空,指针也始终移至else条件。
#include<stdio.h>
#include<malloc.h>
#include<stdlib.h>
#include<string.h>
#include<ctype.h>
#include<stddef.h>
typedef struct node
{
int source;
int num;
struct node *l, *r;
int done;
}node;
void push(int source, int num, struct node *head)
{
node *n = malloc(sizeof(node));
n->num = num;
n->l = NULL;
n->done = 0;
n->source = source;
if (*head == NULL)
{
head = n;
head -> r = NULL;
}
else
{
n -> r = head;
head->l = n;
head = n;
}
}
int pop(node *head)
{
if(head == NULL)
{
return -1;
}
else
{
node *temp = head;
head = head->r;
int num = temp->num;
free(temp);
return num;
}
}
void append(node *extra, node *head)
{
node *temp = extra;
while(temp->r != NULL)
{
temp = temp->r;
}
temp->r = head;
head->l = temp;
head = extra;
}
node** read(int num)
{
char a[2000] = "Assignment1.txt" ,c[1000];
FILE *f = fopen(a,"r");
printf("Got file\n");
node *adj[num];
int i=0;
node *l;
printf("l: %d\n", l);
while(fscanf(f,"%s",c))
{
char *p = strtok(c, ",");
while(p!=NULL)
{
push(i, atoi(p), l);
p = strtok (NULL, ",");
}
adj[i++] = l;
}
printf("Adjacency list created\n");
return adj;
}
node* euler(node *adj[],int n, int i)
{
node *cpath = NULL;
node *fin = NULL;
node *extra;
node *temp = adj[i];
node *tempi;
while(temp!=NULL)
{
if(temp->r->r == NULL)
{
tempi = temp;
}
if(temp->done == 0)
{
temp->done = 1;
push(i, temp->num, cpath);
extra = euler(adj, n, temp->num);
append(extra, cpath);
}
else
{
temp = temp->r;
}
}
while(tempi->l != NULL)
{
push(i,tempi->num, fin);
extra = euler(adj, n, tempi->num);
append(tempi, fin);
tempi = tempi->l;
}
if(tempi != NULL)
{
push(i,tempi->num, fin);
extra = euler(adj, n, tempi->num);
append(tempi, fin);
}
return fin;
}
int main()
{
int n;
printf("Enter the number of vertices: ");
scanf("%d", &n);
node **adj = read(n);
node *fin = euler(adj, n, 0);
node *temp = fin;
while(temp!=NULL)
{
printf("%d ", temp->num);
temp = temp->r;
}
return 0;
}
我尚未调试整个代码,但是我被卡在read()函数中,其中输入是Assignment1.txt,其中包括:
2,3
3,1
1,2
我无法理解为什么出现细分错误。
该函数处理传递给它的指向头节点的指针的值的副本。因此,原始指针本身不会在函数中更改。它是在函数中更改的传递的指针的值的副本。
您需要按引用传递指针,该引用是间接通过指针传递给指针的。
可以通过以下方式声明和定义函数。
int push( struct node **head, int source, int num )
{
node *n = malloc(sizeof(node));
int success = n != NULL;
if ( success )
{
n->source = source;
n->num = num;
n->done = 0;
n->l = NULL;
n->r = *head;
if ( *head != NULL ) ( *head )->l = n;
*head = n;
}
return success;
}
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