怎样实现动态二维数组

[y316237781]

大家好！
我在网上看到一个题：输入一个数，输出矩阵的结果是：
1 3 4 10 11
2 5 9 12 19
6 8 13 18 20
7 14 17 21 24
15 16 22 23 25

这是输入五得到的结果
我现能实现这个矩阵。
我是用的一个二维数组存那些数。
现在我想实现 当我输入一个N，就可以分配一个动态N * N的二维数组。
现在不知怎样实现，请各位指教。
麻烦写具体点。谢谢！

[fallening]

int N;
cin >> N;
int arr[N][N];

//c99 support

[cxxer]

Windows XP SP2, Dev C++。

#include <iostream>
#include <cstdlib>

using namespace std;

template<typename T> class Array;
template<typename T> class ArrayBody
{
    friend class Array<T>;
    T * tpBody;
    int iRows,iColumns, iCurrentRow;
    ArrayBody(int iRsz, int iCsz)
    {
        tpBody = new T[iRsz * iCsz];
        iRows = iRsz; iColumns = iCsz; iCurrentRow = -1;
    }
  public:
    T& operator[](int j)
    {
       bool row_error, column_error;
       row_error = column_error =false;
       try
       {
          if (iCurrentRow < 0 || iCurrentRow >= iRows)
            row_error = true;
          if (row_error == true || column_error == true)
               throw 'e';
       }
       catch(char)
       {
           if (row_error == true)
              cerr << "行下标越界[" << iCurrentRow << "]";
           if (column_error == true)
              cerr << "列下标越界[" << j << "]";
           cout << "\n";
       }
       return tpBody[iCurrentRow * iColumns + j];
    }
    ~ArrayBody() { delete[] tpBody;}
};

template<typename T> class Array
{
      ArrayBody<T> tBody;
  public:
      ArrayBody<T>& operator[](int i)
      {
         tBody.iCurrentRow = i;
         return tBody;
      }
      Array(int iRsz, int iCsz):tBody(iRsz, iCsz){}
};

int main()
{
    int row, col;
    cout << "row:";
    cin >> row;
    cout << "col:";
    cin >> col;
    Array<int> a1(row, col);
    
    int element;
    for(int i = 0; i < row; i++)
    {
            for (int j = 0; j < col; j++)
            {
                cout << "array[" << i << "][" << j << "] = " ;
                cin >> element;
                a1[i][j] = element;
            }
    }
    
     for(int i = 0; i < row; i++)
    {
            for (int j = 0; j < col; j++)
            {
                element = a1[i][j];
                cout << "array[" << i << "][" << j << "] = " <<element << "\t";
            }
            cout << endl;
    }   
    
    system("PAUSE");
    return 0;
    
}

[hengshan]

std::vector<std::vector<type> > table;

这个结构就可以满足你的要求

[shiyi889]

可用链表

[y316237781]

请详细一点，谢谢。
我不会。

[smartjeck]

看看 还有没有程序


二楼水很深 学习

[wyz007134]

int N;
cin >> N;
int* arr = new int[N * N];

使用的时候用行*列 + 列来调用元素

[yunyun1886358]

#include <iostream>

using namespace std;

int main()
{
    int N;
    cin >> N;
    int** pArray = new int* [N];  
    pArray[0] = new int[N * N]; // 分配连续内存  
  
    // 用pArray[1][0]无法寻址，还需指定下标寻址方式  
    for(int i = 1; i < N; i++)  
    {  
        pArray[i] = pArray[i-1] + N;  
    }

    return 0;
}

[耍宝王]

1.
A (*ga)[n] = new A[m][n];
...
delete []ga;
缺点：n必须是已知
优点：调用直观，连续储存，程序简洁(经过测试，析构函数能正确调用)

2.
A** ga = new A*[m];
for(int i = 0; i < m; i++)
ga[i] = new A[n];
...
for(int i = 0; i < m; i++)
delete []ga[i];
delete []ga;
缺点：非连续储存，程序烦琐，ga为A**类型
优点：调用直观，n可以不是已知

3.
A* ga = new A[m*n];
...
delete []ga;
缺点：调用不够直观
优点：连续储存，n可以不是已知

4.
2的改进版
A** ga = new A*[m];
ga[0] = new A[m*n];
for(int i = 1; i < m; i++)
ga[i] = ga[i-1]+n;
优点:连续存储,n可以不是已知,析构方便,猜想只需delete [] ga;
实际：
delete [] ga[0]; // memory for A.
delete [] ga; // memory for A*.
ga = NULL;

[杨卓荦]

vector< vector<int> > a;

[kiss_rule]

我也写了一个。vc6.0
#include <iostream>

using namespace std;

template <typename T>
void MallocArray (T** & array, int const& row, int const& col) {
array = new T* [row];
for (int i = 0; i<row; ++i) {
array[i] = new T [col];
}
}
template <typename T>
void DeleteArray (T** array, int const& row, int const& col ) {
for (int i=0; i<row; ++i) {
delete [] array[i];
}
delete [] array;
}
int main (void) {
int row, col;
cout << "Row : ";
cin >> row;
cout << "col : ";
cin >> col;
int ** dataArray;
MallocArray(dataArray, row, col);
for (int i=0; i<row; ++i)
for (int j=0; j<col; ++j) {
dataArray[i][j] = i*j;
}
for ( i=0; i<row; ++i)
for (int j=0; j<col; ++j){
cout << "dataArray["<< i << "][" << j << "] = " <<dataArray[i][j] << endl;
}
DeleteArray(dataArray, row, col);
return 0;
}

[djjlove_2008]

int arr[M][M];
arr = new int*[M];

for(int i = 0; i < M; ++i)
arr[i] = new int[M];

[lixung]

int *pArr = (int*)malloc(sizeof(int)*N*N);
or
int *pArr = new int[N*N];