Question

您好我正在尝试创建一个简单的链表优先级队列，其中元素根据其f值排序。当我在插入几个元素后打印队列时，我注意到队列中总是只有一个元素（最近插入的元素）。它不包含其他元素。我不确定我做错了什么。

int insertPriorityQueue(struct queueNode* head, struct randomNode* e)
{

struct queueNode* newNode = (struct queueNode*)malloc(sizeof(struct queueNode));

newNode->element = e;

newNode->next = NULL;

if(head==NULL) 
{
    head = newNode;
}

else if(head->element->f > e->f)    
{
        newNode->next = head;
        head = newNode;
}

else
{        
        struct queueNode* temp1 = head;
        struct queueNode* temp2 = NULL;

        while(temp1!= NULL)
        {
            if (temp1->element->f > e->f)               
        {
                    newNode->next = temp1;
                    temp2->next = newNode;
            }
            temp2 = temp1;
            temp1 = temp1->next;
        }

        if(temp2==NULL) 
        {
        head->next = newNode;
    } 

        else if(temp1==NULL)
    {
        temp2->next = newNode;
    } 
}

 printQueue(head);
 return 1;
}

我用来打印队列的函数是：

void printQueue(struct queueNode* head)
{
    struct queueNode* temp = head;
    printf("\n Queue: ");
    while(temp!=NULL)
    {
    printf("[%0.2f]”,temp->element->f);
        temp = temp->next;
}
}

Answer 1

代码评估：

// It is appreciated by answerers if question code is a complete
// program that can be compiled and executed to demonstrate the
// problem. In this case, I will assemble a complete program as 
// the answer...
#include <stdio.h>  // printf();
#include <stdlib.h> // malloc();
#include <string.h> // memset();
#include <errno.h>  // errno

// In the future, the question should provide referenced structures.
// I'll guess at it for demonstration purposes:
typedef struct randomNode
   {
   double f;
   } randomNode_t;

typedef struct queueNode
   {
   struct queueNode  *next;
   randomNode_t      *element;
   } queueNode_t;

void printQueue(struct queueNode* head)
   {
   struct queueNode* temp = head;

   printf("Queue: ");
   while(temp!=NULL)
      {
      printf("[%0.2f]",temp->element->f);
      temp = temp->next;
      }
   printf("\n");   
   }

//int insertPriorityQueue(struct queueNode* head, struct randomNode* e)
//   In order for this function to be able to change the head,
//   it must be supplied the address of head when called.
//   Example: rc = insertPriorityQueue(&head,...
//   Hence, a chage is required to the function definition:
int insertPriorityQueue(struct queueNode **head, struct randomNode *e)
   {
// struct queueNode* newNode = (struct queueNode*)malloc(sizeof(struct queueNode));
//    It is not nessisary to cast the void pointer value returned by malloc().
   struct queueNode *newNode = malloc(sizeof(struct queueNode));
   struct queueNode* temp1 = *head;
   struct queueNode* temp2 = NULL;

// Always check to see if malloc() failed.
   if(!newNode)
      {
      fprintf(stderr, "malloc() reports: %d %s\n", errno, strerror(errno));
      goto CLEANUP;
      }

//    The malloc() function does not "zero-out" the allocated memory,
//    so the following line will wipe the entire structure to zeros.
   memset(newNode, 0, sizeof(*newNode));

//    Flaw here if e changes from each call to this function.   
// newNode->element = e;
//    This is better...
   newNode->element = malloc(sizeof(newNode->element));
   if(!newNode->element)
      {
      fprintf(stderr, "malloc() reports: %d %s\n", errno, strerror(errno));
      goto CLEANUP;
      }

   memcpy(newNode->element, e, sizeof(newNode->element));

// newNode->next is already NULL, due to the above memset().
// newNode->next = NULL;

   /* Check for empty list. */
// if(head==NULL)
//     Since head is now a pointer to a pointer, I would change this line to:
   if(*head == NULL)
      {
//    head = newNode;
//       Now, given you have a pointer to the head value (outside this function)...
//       you can actually change where the head is pointing.
      *head = newNode;
//       Now that there is nothing more to do, just bug-out.
      goto CLEANUP;
      }

   /* Check for head replacement. */
// else if(head->element->f > e->f)
//    Now you can simplify the code by removeing the "else" statement.
//    ...and don't forget to dereference **head...
   if((*head)->element->f > e->f)
      {
      newNode->next = *head;
      *head = newNode;
//       Again, here you are finished.  Remove some complexity by just bugging-out.
      goto CLEANUP;
      }

   /* Determine where this node belongs in the list, and place it there. */
//   Now you can further simplify the code by removing this "else" statement.      
// else
   while(temp1!= NULL)
      {
      if(temp1->element->f > e->f)
         {
         newNode->next = temp1;
         temp2->next = newNode;
// Flaw in question code; should have "bugged-out" here.         
         goto CLEANUP;         
         }

      temp2 = temp1;
      temp1 = temp1->next;
      }

   if(temp2==NULL)
      {
      (*head)->next = newNode;
// Just bug-out, and eliminate the else below.
      goto CLEANUP;
      }

   if(temp1==NULL)
      {
      temp2->next = newNode;
      goto CLEANUP;
      }

CLEANUP:

   printQueue(*head);
   return 1;
   }

int main(void)
   {
   int rCode = 0;
   queueNode_t *head = NULL;
   randomNode_t e;

   e.f = 8;
   insertPriorityQueue(&head, &e);

   e.f = 5;
   insertPriorityQueue(&head, &e);

   e.f = 4;
   insertPriorityQueue(&head, &e);

   e.f = 3;
   insertPriorityQueue(&head, &e);

   e.f = 7;
   insertPriorityQueue(&head, &e);

   e.f = 6;
   insertPriorityQueue(&head, &e);

   return(rCode);   
   }

上面的代码产生以下输出：

Queue: [8.00]
Queue: [5.00][8.00]
Queue: [4.00][5.00][8.00]
Queue: [3.00][4.00][5.00][8.00]
Queue: [3.00][4.00][5.00][7.00][8.00]
Queue: [3.00][4.00][5.00][6.00][7.00][8.00]

C中不正确地插入队列

1 个答案: