矩阵对角线的镜像 · GeeksForGeeks 中文系列教程

# 矩阵对角线的镜像 > 原文： [https://www.geeksforgeeks.org/mirror-matrix-across-diagonal/](https://www.geeksforgeeks.org/mirror-matrix-across-diagonal/) 给定 2 阶 N x N 的二维数组，请打印一个矩阵，该矩阵是对角线上给定树的镜像。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对角线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。 **示例**： ``` 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 ``` **此问题的简单解决方案**占用了额外的空间，我们一对一地遍历所有对角线（从右到左）。在对角线遍历期间，首先将所有元素推入堆栈，然后再次遍历，然后将对角线的每个元素替换为堆栈元素。以下是上述想法的实现。 ## C++ ```cpp // Simple CPP program to find mirror of // matrix across diagonal. #include <bits/stdc++.h> 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<int> 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<int> 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; } ``` ## Java ```java // Simple Java program to find mirror of // matrix across diagonal. import java.util.*; class GFG { static int MAX = 100; static void imageSwap(int mat[][], 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<Integer> s = new Stack<>(); 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.peek(); 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<Integer> s = new Stack<>(); 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.peek(); s.pop(); } } } // Utility function to print a matrix static void printMatrix(int mat[][], int n) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { System.out.print(mat[i][j] + " "); } System.out.println(""); } } // Driver program to test above function public static void main(String[] args) { int mat[][] = {{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); } } // This code contributed by Rajput-Ji ``` ## Python3 ```py # Simple Python3 program to find mirror of # matrix across diagonal. MAX = 100 def imageSwap(mat, n): # for diagonal which start from at # first row of matrix row = 0 # traverse all top right diagonal for j in range(n): # here we use stack for reversing # the element of diagonal s = [] i = row k = j while (i < n and k >= 0): s.append(mat[i][k]) i += 1 k -= 1 # push all element back to matrix # in reverse order i = row k = j while (i < n and k >= 0): mat[i][k] = s[-1] k -= 1 i += 1 s.pop() # do the same process for all the # diagonal which start from last # column column = n - 1 for j in range(1, n): # here we use stack for reversing # the elements of diagonal s = [] i = j k = column while (i < n and k >= 0): s.append(mat[i][k]) i += 1 k -= 1 # push all element back to matrix # in reverse order i = j k = column while (i < n and k >= 0): mat[i][k] = s[-1] i += 1 k -= 1 s.pop() # Utility function to pra matrix def printMatrix(mat, n): for i in range(n): for j in range(n): print(mat[i][j], end=" ") print() # Driver code mat = [[1, 2, 3, 4],[5, 6, 7, 8], [9, 10, 11, 12],[13, 14, 15, 16]] n = 4 imageSwap(mat, n) printMatrix(mat, n) # This code is contributed by shubhamsingh10 ``` ## C# ```cs // Simple C# program to find mirror of // matrix across diagonal. using System; using System.Collections.Generic; class GFG { static int MAX = 100; static void imageSwap(int [,]mat, 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<int> s = new Stack<int>(); 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.Peek(); 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<int> s = new Stack<int>(); 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.Peek(); s.Pop(); } } } // Utility function to print a matrix static void printMatrix(int [,]mat, int n) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { Console.Write(mat[i,j] + " "); } Console.WriteLine(""); } } // Driver code public static void Main(String[] args) { int [,]mat = {{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); } } /* This code contributed by PrinciRaj1992 */ ``` **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） **此问题的有效解决方案**是，如果我们观察到输出矩阵，则注意到我们只需要交换（mat [i] [j] mat [j] [i]）。以下是上述想法的实现。\ ## C++ ``` // Efficient CPP program to find mirror of // matrix across diagonal. #include <bits/stdc++.h> 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; } ```