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!