无法创建链表矩阵

Question

我正在尝试将数组转换为链表矩阵。如果我的输入是：

1 2 3
4 5 6
7 8 9

那么链表矩阵就是这样的：

1 -> 2 -> 3 -> NULL
|    |    |
v    v    v
4 -> 5 -> 6 -> NULL
|    |    |
v    v    v
7 -> 8 -> 9 -> NULL
|    |    |
v    v    v
NULL NULL NULL

我已尝试调试代码。我在col = col->down 中遇到分段错误错误。但我无法理解这个错误背后的原因。这是供您参考的代码。

#include<stdio.h>
#include<stdlib.h>
typedef struct node
{
  int data;
  struct node *next;
  struct node *down;
}matrix;
matrix *start = NULL;
void insert();
void arrToMatrix();
void arrDisplay();
void matrixDisplay();
int a[10][10];
int r,c;
int main()
{
  int n;
  printf("1:Insert elements 2:Convert Array to Linked Matrix 3:Display Array 4:Display Linked Matrix\n");
 for(;;)
 {
   printf("Enter choice: ");
   scanf("%d",&n);
   switch(n)
   {
      case 1: insert();
      break;
      case 2: arrToMatrix();
      break;
      case 3: arrDisplay();
      break;
      case 4: matrixDisplay();
      break;
      default: printf("Wrong Input\n");
      exit(0);
   }
 }
 return 0;
}

void insert()
{
  int i,j;
  printf("Enter row: ");
  scanf("%d",&r);
  printf("\nEnter column: ");
  scanf("%d",&c);
  printf("Enter elements: \n");
  for(i=0;i<r;i++)
      for(j=0;j<c;j++)
        scanf("%d",&a[i][j]);
}
void arrDisplay()
{
  int i,j;
  printf("Array elements: \n");
  for(i=0;i<r;i++)
  {
    for(j=0;j<c;j++)
    {
      printf("%d ",a[i][j]);
    }
    printf("\n");
  }
}
void arrToMatrix()
{
  int i,j;
  matrix *ptr, *col, *row;
  ptr = malloc(sizeof(matrix));
  ptr->data = a[0][0];
  ptr->next = NULL;
  ptr->down = NULL;
  start = ptr;
  col = start;
  for(i=0;i<r;i++)
  {
    row = col;
    for(j=0;j<c;j++)
    {
      ptr = malloc(sizeof(matrix));
      ptr->data = a[i][j];
      ptr->next = NULL;
      ptr->down = NULL;
      if(row == col)
        row = ptr;
      else
      {
        while(row->next!=NULL)
          row = row->next;
        row->next = ptr;
      }
    }
    col = col->down;
  }
}

void matrixDisplay()
{
  matrix *row, *col, *ptr;
  col = start;
  while(col!=NULL)
  {
    row = col;
    while(row!=NULL)
    {
      printf("%d ",row->data);
      row = row->next;
    }
    printf("\n");
    col = col->down;
  }
}

Answer 1

很好的练习。我有一个使用静态指针矩阵方法的解决方案。您可以以相同的方式动态分配它，但只是为了了解一下。

那么，我们开始吧：

#include<stdio.h>
#include<stdlib.h>
typedef struct node
{
  int data;
  struct node *next;
  struct node *down;
} matrix;

matrix *start = NULL;

#define MAX_DIM 10

matrix ptrMatrix[MAX_DIM][MAX_DIM];

void insert();
void arrToMatrix();
void arrDisplay();
void matrixDisplay();

int a[MAX_DIM][MAX_DIM];
int r,c;

int main()
{
  int n;
 for(;;)
 {
   printf("1:Insert elements 2:Convert Array to Linked Matrix 3:Display Array 4:Display Linked Matrix\n");
   printf("Enter choice: ");
   scanf("%d",&n);
   switch(n)
   {
      case 1: insert();
      break;
      case 2: arrToMatrix();
      break;
      case 3: arrDisplay();
      break;
      case 4: matrixDisplay();
      break;
      default: printf("Wrong Input\n");
      exit(0);
   }
 }
 return 0;
}

void insert()
{
  int i,j;
  printf("Enter number of rows: ");
  scanf("%d",&r);
  printf("\nEnter number of columns: ");
  scanf("%d",&c);

  if(r >= MAX_DIM)
  {
      printf("\nNo more than %d rows please!\n");
      return;
  }

  if(c >= MAX_DIM)
  {
      printf("\nNo more than %d columns please!\n");
      return;
  }

  printf("\n Now enter the elements (%d number in total): \n", r*c);
  for(i=0;i<r;i++)
  {
      for(j=0;j<c;j++)
      {
        scanf("%d",&a[i][j]);
      }
  }
}
void arrDisplay()
{
  int i,j;
  printf("Array elements: \n");
  for(i=0;i<r;i++)
  {
    for(j=0;j<c;j++)
    {
      printf("%d ",a[i][j]);
    }
    printf("\n");
  }
}
void arrToMatrix()
{
  if(r<=0 || c <=0)
  {
      printf("Matrix is empty. Try again!\n");
      return;
  }

  int i,j;

  for(i=0;i<r-1;i++)
  {
      for(j=0;j<c-1;j++)
      {
        ptrMatrix[i][j].data = a[i][j];
        ptrMatrix[i][j].next = &ptrMatrix[i][j+1];
        ptrMatrix[i][j].down = &ptrMatrix[i+1][j];
      }

      ptrMatrix[i][c-1].data = a[i][c-1];
      ptrMatrix[i][c-1].next = NULL;
      ptrMatrix[i][c-1].down = &ptrMatrix[i+1][c-1];
  }

  for(j=0;j<c-1;j++)
  {
    ptrMatrix[r-1][j].data = a[r-1][j];
    ptrMatrix[r-1][j].next = &ptrMatrix[r-1][j+1];
    ptrMatrix[r-1][j].down = NULL;
  }

  ptrMatrix[r-1][c-1].data = a[r-1][c-1];
  ptrMatrix[r-1][c-1].next = NULL;
  ptrMatrix[r-1][c-1].down = NULL;

  start = &(ptrMatrix[0][0]);
}

void matrixDisplay()
{
  matrix *row, *col;
  col = start;
  while(col!=NULL)
  {
    row = col;
    while(row!=NULL)
    {
      printf("%d ",row->data);
      row = row->next;
    }
    printf("\n");
    col = col->down;
  }
}

Answer 2

ptr->down 在您的情况下始终为 NULL。它永远不会被初始化。所以 col=col->down 将值 NULL 分配给 col

Answer 3

作业对初学者来说并不容易。

对于初学者来说，使用像start 这样的全局变量并且函数定义依赖于全局变量是个坏主意。

例如，您不能在程序中定义多个矩阵。

在函数arrToMatrix 中，您必须处理指向指针的指针。否则函数会更复杂或者指针 start 不会改变，因为在你的程序中改变局部变量 col 和 row 不会影响 i=on 指针 start。此外，例如在此代码中 sn-p

  ptr = malloc(sizeof(matrix));
  ptr->data = a[i][j];
  ptr->next = NULL;
  ptr->down = NULL;
  if(row == col)
    row = ptr;

当i等于0，j等于0时，存在一个节点的冗余分配。

在函数的开始，由头指针指向的当前矩阵应该被释放。否则可能会出现内存泄漏，用户将无法使用不同的数组重新分配矩阵。

所以你需要再写一个函数来释放当前矩阵的所有分配内存。

这是一个演示程序，展示了如何实现这些功能。在程序中有注释函数matrixDisplay 用于测试。您可以运行它而不是未注释的函数 matrixDisplay 来查看列表及其链接是如何定义的。

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

typedef struct node
{
    int data;
    struct node *next;
    struct node *down;
} matrix;

void deleteMatrix( matrix **head )
{
    if ( *head != NULL )
    {
        for ( matrix *row = ( *head )->next; row != NULL; )
        {
            while ( row != NULL )
            {
                matrix *tmp = row;
                row = row->next;
                free( tmp );
            }

            row = ( *head )->down;

            if ( ( *head )->down != NULL )
            {
                ( *head )->down = row->down;
            }
        }

        free( *head );
        *head = NULL;
    }
}

void arrToMatrix( matrix **head, size_t m, size_t n, int a[m][n] )
{
    deleteMatrix( head );

    matrix **row = head;

    for ( size_t i = 0; i < m; i++ )
    {
        matrix **column = row;

        for ( size_t j = 0; j < n; j++ )
        {
            *column = malloc( sizeof( matrix ) );
            ( *column )->data = a[i][j];
            ( *column )->down = NULL;
            ( *column )->next = NULL;

            column = &( *column )->next;    
        }

        row = &( *row )->down;
    }

    row = head;

    for ( size_t i = 0; i + 1 < m; i++ )
    {
        matrix *column   = *row;
        matrix *next_row_column = ( *row )->down;

        for ( size_t j = 0; j < n; j++ )
        {
            column->down = next_row_column;
            column = column->next;
            next_row_column = next_row_column->next;
        }           

        row = &( *row )->down;
    }
}

void matrixDisplay( const matrix *head )
{
    for ( ; head != NULL; head = head->down )
    {
        for ( const matrix *column = head; column != NULL; column = column->next )
        {
            printf( "%2d ", column->data );
        }
        putchar( '\n' );
    }
}
/*
void matrixDisplay( const matrix *head )
{
    for ( ; head != NULL; head = head->down )
    {
        for ( const matrix *column = head; column != NULL; column = column->next )
        {
            printf( "%2d ( %p ): next( %p ), down( %p )  ", 
                     column->data, ( void * )column, ( void * )column->next, ( void * )column->down );
        }
        putchar( '\n' );
    }
}
*/
int main(void) 
{
    enum { N = 3 };
    int a[][N] = 
    {
        { 1, 2, 3 },
        { 4, 5, 6 },
        { 7, 8, 9 }
    };

    matrix *head = NULL;

    arrToMatrix( &head, N, N, a );

    matrixDisplay( head );
    putchar( '\n' );

    deleteMatrix( &head );

    return 0;
}

程序输出是

 1  2  3 
 4  5  6 
 7  8  9 

带注释的函数 matrixDisplay 的输出可能如下所示。

 1 ( 0x558b577e7260 ): next( 0x558b577e7280 ), down( 0x558b577e72c0 )   2 ( 0x558b577e7280 ): next( 0x558b577e72a0 ), down( 0x558b577e72e0 )   3 ( 0x558b577e72a0 ): next( (nil) ), down( 0x558b577e7300 )  
 4 ( 0x558b577e72c0 ): next( 0x558b577e72e0 ), down( 0x558b577e7320 )   5 ( 0x558b577e72e0 ): next( 0x558b577e7300 ), down( 0x558b577e7340 )   6 ( 0x558b577e7300 ): next( (nil) ), down( 0x558b577e7360 )  
 7 ( 0x558b577e7320 ): next( 0x558b577e7340 ), down( (nil) )   8 ( 0x558b577e7340 ): next( 0x558b577e7360 ), down( (nil) )   9 ( 0x558b577e7360 ): next( (nil) ), down( (nil) )