给定一个nxn阶的二维数组，在对角线上打印一个矩阵，它是给定树的镜像。我们需要以一种方式打印结果：将对角线上方三角形的值与其下方三角形的值进行交换，就像镜像交换一样。打印在矩阵布局中获得的二维阵列。
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例如：
Input : int mat[][] = {{1 2 4 } {5 9 0} { 3 1 7}}Output : 1 5 3 2 9 1 4 0 7Input : 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
A. 简单解决方案 解决这个问题需要额外的空间。我们一个接一个地穿过所有右对角线（从右到左）。在遍历对角线的过程中，首先，我们将所有元素推入堆栈，然后再次遍历它，并用堆栈元素替换对角线的每个元素。
下面是上述想法的实施。
C++
// 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
// 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
# 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#
// 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 */ |
Javascript
<script> // Simple Javascript program to find mirror of matrix across diagonal. let MAX = 100; function imageSwap(mat, n) { // for diagonal which start from at // first row of matrix let row = 0; // traverse all top right diagonal for (let j = 0; j < n; j++) { // here we use stack for reversing // the element of diagonal let s = []; let 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[s.length - 1]; s.pop(); } } // do the same process for all the // diagonal which start from last // column let column = n - 1; for (let j = 1; j < n; j++) { // here we use stack for reversing // the elements of diagonal let s = []; let 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[s.length - 1]; s.pop(); } } } // Utility function to print a matrix function printMatrix(mat, n) { for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { document.write(mat[i][j] + " " ); } document.write( "</br>" ); } } let mat = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]; let n = 4; imageSwap(mat, n); printMatrix(mat, n); </script> |
输出：
1 5 9 13 2 6 10 14 3 7 11 15 4 8 12 16
时间复杂性： O（n*n）
一 有效解决方案 对于这个问题，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]）。 下面是上述想法的实施。
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; } |
JAVA
// Efficient Java program to find mirror of // matrix across diagonal. import java.io.*; class GFG { static int MAX = 100 ; static void imageSwap( int mat[][], 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 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 is contributed by anuj_67. |
Python3
# Efficient Python3 program to find mirror of # matrix across diagonal. from builtins import range MAX = 100 ; def imageSwap(mat, n): # traverse a matrix and swap # mat[i][j] with mat[j][i] for i in range (n): for j in range (i + 1 ): t = mat[i][j]; mat[i][j] = mat[j][i] mat[j][i] = t # 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 if __name__ = = '__main__' : 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 Rajput-Ji |
C#
// Efficient C# program to find mirror of // matrix across diagonal. using System; class GFG { //static int MAX = 100; static void imageSwap( int [,]mat, 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 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 program to test above function public static void Main () { 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 is contributed by anuj_67. |
PHP
<?php // Efficient PHP program to find mirror // of matrix across diagonal. function imageSwap(& $mat , $n ) { // traverse a matrix and swap // mat[i][j] with mat[j][i] for ( $i = 0; $i < $n ; $i ++) for ( $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 function printMatrix(& $mat , $n ) { for ( $i = 0; $i < $n ; $i ++) { for ( $j = 0; $j < $n ; $j ++) { echo ( $mat [ $i ][ $j ]); echo ( " " ); } echo ( "" ); } } // Driver Code $mat = array ( array (1, 2, 3, 4), array (5, 6, 7, 8), array (9, 10, 11, 12), array (13, 14, 15, 16)); $n = 4; imageSwap( $mat , $n ); printMatrix( $mat , $n ); // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script> // Efficient Javascript program to find mirror of // matrix across diagonal. let MAX = 100; function imageSwap(mat, n) { // traverse a matrix and swap // mat[i][j] with mat[j][i] for (let i = 0; i < n; i++) for (let 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 function printMatrix(mat, n) { for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) document.write(mat[i][j] + " " ); document.write( "</br>" ); } } let mat = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ] ]; let n = 4; imageSwap(mat, n); printMatrix(mat, n); </script> |
输出：
1 5 9 13 2 6 10 14 3 7 11 15 4 8 12 16
时间复杂性： O（n*n）
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