📜  矩阵对角线镜像的C++程序

📅  最后修改于: 2022-05-13 01:55:29.027000             🧑  作者: Mango

矩阵对角线镜像的C++程序

给定一个 N x N 阶的二维数组,打印一个矩阵,它是给定树在对角线上的镜像。我们需要以某种方式打印结果:将对角线上方的三角形的值与其下方的三角形的值交换,就像镜像交换一样。打印以矩阵布局获得的二维数组。

例子:

Input : int mat[][] = {{1 2 4 }
                       {5 9 0}
                       { 3 1 7}}
Output :  1 5 3 
          2 9 1
          4 0 7

Input : mat[][] = {{1  2  3  4 }
                   {5  6  7  8 }
                   {9  10 11 12}
                   {13 14 15 16} }
Output : 1 5 9 13 
         2 6 10 14  
         3 7 11 15 
         4 8 12 16 

这个问题的一个简单解决方案涉及额外的空间。我们一一遍历所有右对角线(从右到左)。在遍历对角线的过程中,首先我们将所有元素压入栈中,然后再次遍历它,并将对角线的每个元素替换为栈元素。

下面是上述思想的实现。

C++
// Simple CPP program to find mirror of
// matrix across diagonal.
#include 
using namespace std;
  
const int MAX = 100;
  
void imageSwap(int mat[][MAX], int n)
{
    // for diagonal which start from at 
    // first row of matrix
    int row = 0;
  
    // traverse all top right diagonal
    for (int j = 0; j < n; j++) {
  
        // here we use stack for reversing
        // the element of diagonal
        stack s;
        int i = row, k = j;
        while (i < n && k >= 0) 
            s.push(mat[i++][k--]);
          
        // push all element back to matrix 
        // in reverse order
        i = row, k = j;
        while (i < n && k >= 0) {
            mat[i++][k--] = s.top();
            s.pop();
        }
    }
  
    // do the same process for all the
    // diagonal which start from last
    // column
    int column = n - 1;
    for (int j = 1; j < n; j++) {
  
        // here we use stack for reversing 
        // the elements of diagonal
        stack s;
        int i = j, k = column;
        while (i < n && k >= 0) 
            s.push(mat[i++][k--]);
          
        // push all element back to matrix 
        // in reverse order
        i = j;
        k = column;
        while (i < n && k >= 0) {
            mat[i++][k--] = s.top();
            s.pop();
        }
    }
}
  
// Utility function to print a matrix
void printMatrix(int mat[][MAX], int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cout << mat[i][j] << " ";
        cout << endl;
    }
}
  
// driver program to test above function
int main()
{
    int mat[][MAX] = { { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 },
                     { 9, 10, 11, 12 },
                     { 13, 14, 15, 16 } };
    int n = 4;
    imageSwap(mat, n);
    printMatrix(mat, n);
    return 0;
}


C++
// Efficient CPP program to find mirror of
// matrix across diagonal.
#include 
using namespace std;
  
const int MAX = 100;
  
void imageSwap(int mat[][MAX], int n)
{
    // traverse a matrix and swap 
    // mat[i][j] with mat[j][i]
    for (int i = 0; i < n; i++)
        for (int j = 0; j <= i; j++) 
            mat[i][j] = mat[i][j] + mat[j][i] - 
                       (mat[j][i] = mat[i][j]);       
}
  
// Utility function to print a matrix
void printMatrix(int mat[][MAX], int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cout << mat[i][j] << " ";
        cout << endl;
    }
}
  
// driver program to test above function
int main()
{
    int mat[][MAX] = { { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 },
                     { 9, 10, 11, 12 },
                     { 13, 14, 15, 16 } };
    int n = 4;
    imageSwap(mat, n);
    printMatrix(mat, n);
    return 0;
}


输出:

1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16

时间复杂度: O(n*n)

这个问题的一个有效解决方案是,如果我们观察一个输出矩阵,那么我们注意到我们只需要交换 (mat[i][j] 到 mat[j][i])。
下面是上述思想的实现。

C++

// Efficient CPP program to find mirror of
// matrix across diagonal.
#include 
using namespace std;
  
const int MAX = 100;
  
void imageSwap(int mat[][MAX], int n)
{
    // traverse a matrix and swap 
    // mat[i][j] with mat[j][i]
    for (int i = 0; i < n; i++)
        for (int j = 0; j <= i; j++) 
            mat[i][j] = mat[i][j] + mat[j][i] - 
                       (mat[j][i] = mat[i][j]);       
}
  
// Utility function to print a matrix
void printMatrix(int mat[][MAX], int n)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cout << mat[i][j] << " ";
        cout << endl;
    }
}
  
// driver program to test above function
int main()
{
    int mat[][MAX] = { { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 },
                     { 9, 10, 11, 12 },
                     { 13, 14, 15, 16 } };
    int n = 4;
    imageSwap(mat, n);
    printMatrix(mat, n);
    return 0;
}

输出:

1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16 

时间复杂度: O(n*n)

有关详细信息,请参阅关于矩阵对角线镜像的完整文章!