给定2D矩阵，按对角线顺序打印给定矩阵的所有元素。例如，考虑下面的5×4输入矩阵。
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例子：
1 2 3 45 6 7 89 10 11 1213 14 15 1617 18 19 20
上述矩阵的对角线打印为
15 29 6 313 10 7 417 14 11 818 15 1219 1620
另一个例子 :
我们强烈建议您在继续解决方案之前单击此处并进行练习。
下面是对角线打印的代码。 给定矩阵的对角线打印“矩阵[行][COL]”输出中始终有“行+列–1”行。
C++
// C++ program to print all elements // of given matrix in diagonal order #include <bits/stdc++.h> using namespace std; #define ROW 5 #define COL 4 // A utility function to find min // of two integers int minu( int a, int b) { return (a < b) ? a : b; } // A utility function to find min // of three integers int min( int a, int b, int c) { return minu(minu(a, b), c); } // A utility function to find // max of two integers int max( int a, int b) { return (a > b) ? a : b; } // The main function that prints given // matrix in diagonal order void diagonalOrder( int matrix[][COL]) { // There will be ROW+COL-1 lines // in the output for ( int line = 1; line <= (ROW + COL - 1); line++) { /* Get column index of the first element in this line of output. The index is 0 for first ROW lines and line - ROW for remaining lines */ int start_col = max(0, line - ROW); /* Get count of elements in this line. The count of elements is equal to minimum of line number, COL-start_col and ROW */ int count = min(line, (COL - start_col), ROW); /* Print elements of this line */ for ( int j = 0; j < count; j++) cout << setw(5) << matrix[minu(ROW, line) - j - 1][start_col + j]; /* Print elements of next diagonal on next line */ cout << "" ; } } // Utility function to print a matrix void printMatrix( int matrix[ROW][COL]) { for ( int i = 0; i < ROW; i++) { for ( int j = 0; j < COL; j++) cout << setw(5) << matrix[i][j]; cout << "" ; } } // Driver code int main() { int M[ROW][COL] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 }, { 17, 18, 19, 20 },}; cout << "Given matrix is " << endl; printMatrix(M); cout << "Diagonal printing of matrix is " << endl; diagonalOrder(M); return 0; } // This code is contributed by shubhamsingh10 |
C
// C program to print all elements // of given matrix in diagonal order #include <stdlib.h> #define ROW 5 #define COL 4 // A utility function to find min of two integers int minu( int a, int b) { return (a < b)? a: b; } // A utility function to find min of three integers int min( int a, int b, int c) { return minu(minu(a, b), c);} // A utility function to find max of two integers int max( int a, int b) { return (a > b)? a: b; } // The main function that prints given matrix in // diagonal order void diagonalOrder( int matrix[][COL]) { // There will be ROW+COL-1 lines in the output for ( int line=1; line<=(ROW + COL -1); line++) { /* Get column index of the first element in this line of output. The index is 0 for first ROW lines and line - ROW for remaining lines */ int start_col = max(0, line-ROW); /* Get count of elements in this line. The count of elements is equal to minimum of line number, COL-start_col and ROW */ int count = min(line, (COL-start_col), ROW); /* Print elements of this line */ for ( int j=0; j<count; j++) printf ( "%5d " , matrix[minu(ROW, line)-j-1][start_col+j]); /* Print elements of next diagonal on next line */ printf ( "" ); } } // Utility function to print a matrix void printMatrix( int matrix[ROW][COL]) { for ( int i=0; i< ROW; i++) { for ( int j=0; j<COL; j++) printf ( "%5d " , matrix[i][j]); printf ( "" ); } } // Driver code int main() { int M[ROW][COL] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}, {13, 14, 15, 16}, {17, 18, 19, 20}, }; printf ( "Given matrix is " ); printMatrix(M); printf ( "Diagonal printing of matrix is " ); diagonalOrder(M); return 0; } |
JAVA
// Java program to print all elements // of given matrix in diagonal order class GFG { static final int ROW = 5 ; static final int COL = 4 ; // A utility function to find min // of two integers static int min( int a, int b) { return (a < b) ? a : b; } // A utility function to find min // of three integers static int min( int a, int b, int c) { return min(min(a, b), c); } // A utility function to find max // of two integers static int max( int a, int b) { return (a > b) ? a : b; } // The main function that prints given // matrix in diagonal order static void diagonalOrder( int matrix[][]) { // There will be ROW+COL-1 lines in the output for ( int line = 1 ; line <= (ROW + COL - 1 ); line++) { // Get column index of the first // element in this line of output. // The index is 0 for first ROW // lines and line - ROW for remaining lines int start_col = max( 0 , line - ROW); // Get count of elements in this line. // The count of elements is equal to // minimum of line number, COL-start_col and ROW int count = min(line, (COL - start_col), ROW); // Print elements of this line for ( int j = 0 ; j < count; j++) System.out.print(matrix[min(ROW, line) - j- 1 ][start_col + j] + " " ); // Print elements of next diagonal on next line System.out.println(); } } // Utility function to print a matrix static void printMatrix( int matrix[][]) { for ( int i = 0 ; i < ROW; i++) { for ( int j = 0 ; j < COL; j++) System.out.print(matrix[i][j] + " " ); System.out.print( "" ); } } // Driver code public static void main(String[] args) { int M[][] = { { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 10 , 11 , 12 }, { 13 , 14 , 15 , 16 }, { 17 , 18 , 19 , 20 }, }; System.out.print( "Given matrix is " ); printMatrix(M); System.out.print( "Diagonal printing of matrix is " ); diagonalOrder(M); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 program to print all elements # of given matrix in diagonal order ROW = 5 COL = 4 # Main function that prints given # matrix in diagonal order def diagonalOrder(matrix): # There will be ROW+COL-1 lines in the output for line in range ( 1 , (ROW + COL)): # Get column index of the first element # in this line of output. The index is 0 # for first ROW lines and line - ROW for # remaining lines start_col = max ( 0 , line - ROW) # Get count of elements in this line. # The count of elements is equal to # minimum of line number, COL-start_col and ROW count = min (line, (COL - start_col), ROW) # Print elements of this line for j in range ( 0 , count): print (matrix[ min (ROW, line) - j - 1 ] [start_col + j], end = " " ) print () # Utility function to print a matrix def printMatrix(matrix): for i in range ( 0 , ROW): for j in range ( 0 , COL): print (matrix[i][j], end = " " ) print () # Driver Code M = [[ 1 , 2 , 3 , 4 ], [ 5 , 6 , 7 , 8 ], [ 9 , 10 , 11 , 12 ], [ 13 , 14 , 15 , 16 ], [ 17 , 18 , 19 , 20 ]] print ( "Given matrix is " ) printMatrix(M) print ( "Diagonal printing of matrix is " ) diagonalOrder(M) # This code is contributed by Nikita Tiwari. |
C#
// C# program to print all elements // of given matrix in diagonal order using System; using static System.Math; class GFG { static int ROW = 5; static int COL = 4; // The main function that prints given // matrix in diagonal order static void diagonalOrder( int [, ] matrix) { // There will be ROW+COL-1 lines in the output for ( int line = 1; line <= (ROW + COL - 1); line++) { // Get column index of the first element // in this line of output.The index is 0 // for first ROW lines and line - ROW for // remaining lines int start_col = Max(0, line - ROW); // Get count of elements in this line. The // count of elements is equal to minimum of // line number, COL-start_col and ROW int count = Min(line, Math.Min((COL - start_col), ROW)); // Print elements of this line for ( int j = 0; j < count; j++) Console.Write(matrix[Min(ROW, line) - j - 1, start_col + j] + " " ); // Print elements of next diagonal // on next line Console.WriteLine(); } } // Utility function to print a matrix static void printMatrix( int [, ] matrix) { for ( int i = 0; i < ROW; i++) { for ( int j = 0; j < COL; j++) Console.Write(matrix[i, j] + " " ); Console.WriteLine( "" ); } } // Driver code public static void Main() { int [, ] M = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 }, { 17, 18, 19, 20 } }; Console.Write( "Given matrix is " ); printMatrix(M); Console.Write( "Diagonal printing" + " of matrix is " ); diagonalOrder(M); } } // This code is contributed by Sam007. |
PHP
<?php // PHP Code for Zigzag (or diagonal) // traversal of Matrix $ROW = 5; $COL = 4; // The main function that prints // given matrix in diagonal order function diagonalOrder(& $matrix ) { global $ROW , $COL ; // There will be ROW+COL-1 // lines in the output for ( $line = 1; $line <= ( $ROW + $COL - 1); $line ++) { /* Get column index of the first element in this line of output. The index is 0 for first ROW lines and line - ROW for remaining lines */ $start_col = max(0, $line - $ROW ); /* Get count of elements in this line. The count of elements is equal to minimum of line number, COL-start_col and ROW */ $count = min( $line , ( $COL - $start_col ), $ROW ); /* Print elements of this line */ for ( $j = 0; $j < $count ; $j ++) { echo $matrix [min( $ROW , $line ) - $j - 1][ $start_col + $j ]; echo " " ; } /* Print elements of next diagonal on next line */ print ( "" ); } } // Utility function // to print a matrix function printMatrix(& $matrix ) { global $ROW , $COL ; for ( $i = 0; $i < $ROW ; $i ++) { for ( $j = 0; $j < $COL ; $j ++) { echo $matrix [ $i ][ $j ] ; echo " " ; } print ( "" ); } } // Driver Code $M = array ( array (1, 2, 3, 4), array (5, 6, 7, 8), array (9, 10, 11, 12), array (13, 14, 15, 16), array (17, 18, 19, 20)); echo "Given matrix is " ; printMatrix( $M ); printf ( "Diagonal printing " . "of matrix is " ); diagonalOrder( $M ); // This code is contributed // by ChitraNayal ?> |
Javascript
<script> // Javascript program to print all elements // of given matrix in diagonal order let ROW = 5; let COL = 4; // A utility function to find min // of two integers function min(a, b) { return (a < b) ? a : b; } // A utility function to find min // of three integer function _min(a, b, c) { return min(min(a, b), c); } // A utility function to find max // of two integers function max(a,b) { return (a > b) ? a : b; } // The main function that prints given // matrix in diagonal order function diagonalOrder(matrix) { // There will be ROW+COL-1 lines in the output for (let line = 1; line <= (ROW + COL - 1); line++) { // Get column index of the first // element in this line of output. // The index is 0 for first ROW // lines and line - ROW for remaining lines let start_col = max(0, line - ROW); // Get count of elements in this line. // The count of elements is equal to // minimum of line number, COL-start_col and ROW let count = min(line, (COL - start_col), ROW); // Print elements of this line for (let j = 0; j < count; j++) document.write(matrix[min(ROW, line) - j- 1][start_col + j] + " " ); // Print elements of next diagonal on next line document.write( "<br>" ); } } // Utility function to print a matrix function printMatrix(matrix) { for (let i = 0; i < ROW; i++) { for (let j = 0; j < COL; j++) document.write(matrix[i][j] + " " ); document.write( "<br>" ); } } // Driver code let M = [[ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ], [ 17, 18, 19, 20 ]]; document.write( "Given matrix is <br>" ); printMatrix(M); document.write( "<br>Diagonal printing of matrix is <br>" ); diagonalOrder(M); // This code is contributed by ab2127 </script> |
输出
Given matrix is 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Diagonal printing of matrix is 1 5 2 9 6 3 13 10 7 4 17 14 11 8 18 15 12 19 16 20
下面是一个例子 替代方法 解决上述问题。
Matrix => 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Observe the sequence 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / 13 / 14 / 15 / 16 / 17 / 18 / 19 / 20
C++
#include <bits/stdc++.h> #define R 5 #define C 4 using namespace std; bool isValid( int i, int j) { if (i < 0 || i >= R || j >= C || j < 0) return false ; return true ; } void diagonalOrder( int arr[][C]) { /* through this for loop we choose each element of first column as starting point and print diagonal starting at it. arr[0][0], arr[1][0]....arr[R-1][0] are all starting points */ for ( int k = 0; k < R; k++) { cout << arr[k][0] << " " ; // set row index for next point in // diagonal int i = k - 1; // set column index for next point in // diagonal int j = 1; /* Print Diagonally upward */ while (isValid(i, j)) { cout << arr[i][j] << " " ; i--; // move in upright direction j++; } cout << endl; } /* through this for loop we choose each element of last row as starting point (except the [0][c-1] it has already been processed in previous for loop) and print diagonal starting at it. arr[R-1][0], arr[R-1][1]....arr[R-1][c-1] are all starting points */ // Note : we start from k = 1 to C-1; for ( int k = 1; k < C; k++) { cout << arr[R - 1][k] << " " ; // set row index for next point in // diagonal int i = R - 2; // set column index for next point in // diagonal int j = k + 1; /* Print Diagonally upward */ while (isValid(i, j)) { cout << arr[i][j] << " " ; i--; // move in upright direction j++; } cout << endl; } } // Driver Code int main() { int arr[][C] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 }, { 17, 18, 19, 20 }, }; diagonalOrder(arr); return 0; } |
JAVA
// JAVA Code for Zigzag (or diagonal) // traversal of Matrix class GFG { public static int R, C; private static void diagonalOrder( int [][] arr) { /* through this for loop we choose each element of first column as starting point and print diagonal starting at it. arr[0][0], arr[1][0]....arr[R-1][0] are all starting points */ for ( int k = 0 ; k < R; k++) { System.out.print(arr[k][ 0 ] + " " ); // set row index for next // point in diagonal int i = k - 1 ; // set column index for // next point in diagonal int j = 1 ; /* Print Diagonally upward */ while (isValid(i, j)) { System.out.print(arr[i][j] + " " ); i--; // move in upright direction j++; } System.out.println( "" ); } /* through this for loop we choose each element of last row as starting point (except the [0][c-1] it has already been processed in previous for loop) and print diagonal starting at it. arr[R-1][0], arr[R-1][1].... arr[R-1][c-1] are all starting points */ // Note : we start from k = 1 to C-1; for ( int k = 1 ; k < C; k++) { System.out.print(arr[R - 1 ][k] + " " ); // set row index for next // point in diagonal int i = R - 2 ; // set column index for // next point in diagonal int j = k + 1 ; /* Print Diagonally upward */ while (isValid(i, j)) { System.out.print(arr[i][j] + " " ); // move in upright direction i--; j++; } System.out.println( "" ); } } public static boolean isValid( int i, int j) { if (i < 0 || i >= R || j >= C || j < 0 ) return false ; return true ; } // Driver code public static void main(String[] args) { int arr[][] = { { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 10 , 11 , 12 }, { 13 , 14 , 15 , 16 }, { 17 , 18 , 19 , 20 }, }; R = arr.length; C = arr[ 0 ].length; // Function call diagonalOrder(arr); } } // This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 program to print all elements # of given matrix in diagonal order R = 5 C = 4 def isValid(i, j): if (i < 0 or i > = R or j > = C or j < 0 ): return False return True def diagonalOrder(arr): # through this for loop we choose each element # of first column as starting point and print # diagonal starting at it. # arr[0][0], arr[1][0]....arr[R-1][0] # are all starting points for k in range ( 0 , R): print (arr[k][ 0 ], end = " " ) # set row index for next point in diagonal i = k - 1 # set column index for next point in diagonal j = 1 # Print Diagonally upward while (isValid(i, j)): print (arr[i][j], end = " " ) i - = 1 j + = 1 # move in upright direction print () # Through this for loop we choose each # element of last row as starting point # (except the [0][c-1] it has already been # processed in previous for loop) and print # diagonal starting at it. # arr[R-1][0], arr[R-1][1]....arr[R-1][c-1] # are all starting points # Note : we start from k = 1 to C-1; for k in range ( 1 , C): print (arr[R - 1 ][k], end = " " ) # set row index for next point in diagonal i = R - 2 # set column index for next point in diagonal j = k + 1 # Print Diagonally upward while (isValid(i, j)): print (arr[i][j], end = " " ) i - = 1 j + = 1 # move in upright direction print () # Driver Code arr = [[ 1 , 2 , 3 , 4 ], [ 5 , 6 , 7 , 8 ], [ 9 , 10 , 11 , 12 ], [ 13 , 14 , 15 , 16 ], [ 17 , 18 , 19 , 20 ]] # Function call diagonalOrder(arr) # This code is contributed by Nikita Tiwari. |
C#
// C# Code for Zigzag (or diagonal) // traversal of Matrix using System; class GFG { public static int R, C; private static void diagonalOrder( int [, ] arr) { /* through this for loop we choose each element of first column as starting point and print diagonal starting at it. arr[0,0], arr[1,0]....arr[R-1,0] are all starting points */ for ( int k = 0; k < R; k++) { Console.Write(arr[k, 0] + " " ); // set row index for next // point in diagonal int i = k - 1; // set column index for // next point in diagonal int j = 1; /* Print Diagonally upward */ while (isValid(i, j)) { Console.Write(arr[i, j] + " " ); i--; // move in upright direction j++; } Console.Write( "" ); } /* through this for loop we choose each element of last row as starting point (except the [0][c-1] it has already been processed in previous for loop) and print diagonal starting at it. arr[R-1,0], arr[R-1,1].... arr[R-1,c-1] are all starting points */ // Note : we start from k = 1 to C-1; for ( int k = 1; k < C; k++) { Console.Write(arr[R - 1, k] + " " ); // set row index for next // point in diagonal int i = R - 2; // set column index for // next point in diagonal int j = k + 1; /* Print Diagonally upward */ while (isValid(i, j)) { Console.Write(arr[i, j] + " " ); i--; j++; // move in upright direction } Console.Write( "" ); } } public static bool isValid( int i, int j) { if (i < 0 || i >= R || j >= C || j < 0) return false ; return true ; } // Driver code public static void Main() { int [, ] arr = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 }, { 17, 18, 19, 20 } }; R = arr.GetLength(0); C = arr.GetLength(1); // Function call diagonalOrder(arr); } } // This code is contributed // by ChitraNayal |
PHP
<?php // PHP code for Zigzag (or diagonal) // traversal of Matrix define( "R" , 5); define( "C" , 4); function isValid( $i , $j ) { if ( $i < 0 || $i >= R || $j >= C || $j < 0) return false; return true; } function diagonalOrder(& $arr ) { /* through this for loop we choose each element of first column as starting point and print diagonal starting at it. arr[0][0], arr[1][0]....arr[R-1][0] are all starting points */ for ( $k = 0; $k < R; $k ++) { echo $arr [ $k ][0] . " " ; $i = $k - 1; // set row index for next // point in diagonal $j = 1; // set column index for next // point in diagonal /* Print Diagonally upward */ while (isValid( $i , $j )) { echo $arr [ $i ][ $j ] . " " ; $i --; $j ++; // move in upright direction } echo "" ; } /* through this for loop we choose each element of last row as starting point (except the [0][c-1] it has already been processed in previous for loop) and print diagonal starting at it. arr[R-1][0], arr[R-1][1]....arr[R-1][c-1] are all starting points */ //Note : we start from k = 1 to C-1; for ( $k = 1; $k < C; $k ++) { echo $arr [R - 1][ $k ] . " " ; $i = R - 2; // set row index for next // point in diagonal $j = $k + 1; // set column index for next // point in diagonal /* Print Diagonally upward */ while (isValid( $i , $j )) { echo $arr [ $i ][ $j ] . " " ; $i --; $j ++; // move in upright direction } echo "" ; } } // Driver Code $arr = array ( array (1, 2, 3, 4), array (5, 6, 7, 8), array (9, 10, 11, 12), array (13, 14, 15, 16), array (17, 18, 19, 20)); // Function call diagonalOrder( $arr ); // This code is contributed // by rathbhupendra ?> |
Javascript
<script> // JAVA Code for Zigzag (or diagonal) // traversal of Matrix var R, C; function diagonalOrder( arr) { /* through this for loop we choose each element of first column as starting point and print diagonal starting at it. arr[0][0], arr[1][0]....arr[R-1][0] are all starting points */ for ( var k = 0; k < R; k++) { document.write(arr[k][0] + " " ); // set row index for next // point in diagonal var i = k - 1; // set column index for // next point in diagonal var j = 1; /* Print Diagonally upward */ while (isValid(i, j)) { document.write(arr[i][j] + " " ); i--; // move in upright direction j++; } document.writeln( "<br>" ); } /* through this for loop we choose each element of last row as starting point (except the [0][c-1] it has already been processed in previous for loop) and print diagonal starting at it. arr[R-1][0], arr[R-1][1].... arr[R-1][c-1] are all starting points */ // Note : we start from k = 1 to C-1; for ( var k = 1; k < C; k++) { document.write(arr[R - 1][k] + " " ); // set row index for next // point in diagonal var i = R - 2; // set column index for // next point in diagonal var j = k + 1; /* Print Diagonally upward */ while (isValid(i, j)) { document.write(arr[i][j] + " " ); // move in upright direction i--; j++; } document.writeln( "<br>" ); } } function isValid( i, j) { if (i < 0 || i >= R || j >= C || j < 0) return false ; return true ; } // Driver code var arr = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ], [ 17, 18, 19, 20 ], ]; R = arr.length; C = arr[0].length; // Function call diagonalOrder(arr); // This code is contributed by shivanisinghss2110 </script> |
输出
1 5 2 9 6 3 13 10 7 4 17 14 11 8 18 15 12 19 16 20
感谢Gaurav Ahirwar提出这种方法。
本文由 阿什阿南德 并由Geeksforgeks团队审核。如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。
另一种方法：
这是一个敏锐的观察，即[i+j]的总和，即数组的索引，在整个对角线上保持不变。因此，我们将利用矩阵的这个特性，使我们的代码简短而简单。
![图片[2]-矩阵的之字形（或对角线）遍历-yiteyi-C++库](https://media.geeksforgeeks.org/wp-content/uploads/20200716201927/matrix.png)
以下是上述理念的实施情况：
C++
#include <bits/stdc++.h> #define R 5 #define C 4 using namespace std; void diagonalOrder( int arr[][C], int n, int m) { // we will use a 2D vector to // store the diagonals of our array // the 2D vector will have (n+m-1) // rows that is equal to the number of // diagonals vector<vector< int > > ans(n + m - 1); for ( int i = 0; i < m; i++) { for ( int j = 0; j < n; j++) { ans[i + j].push_back(arr[j][i]); } } for ( int i = 0; i < ans.size(); i++) { for ( int j = 0; j < ans[i].size(); j++) cout << ans[i][j] << " " ; cout << endl; } } // Driver Code int main() { // we have a matrix of n rows // and m columns int n = 5, m = 4; int arr[][C] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 }, { 17, 18, 19, 20 }, }; // Function call diagonalOrder(arr, n, m); return 0; } |
JAVA
import java.util.*; import java.io.*; class GFG { public static int R = 5 , C = 4 ; public static void diagonalOrder( int [][] arr, int n, int m) { // we will use a 2D vector to // store the diagonals of our array // the 2D vector will have (n+m-1) // rows that is equal to the number of // diagonals ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>(n+m- 1 ); for ( int i = 0 ; i < n + m - 1 ; i++) { ans.add( new ArrayList<Integer>()); } for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < m; j++) { (ans.get(i+j)).add(arr[i][j]); } } for ( int i = 0 ; i < ans.size(); i++) { for ( int j = ans.get(i).size() - 1 ; j >= 0 ; j--) { System.out.print(ans.get(i).get(j)+ " " ); } System.out.println(); } } // Driver code public static void main (String[] args) { int n = 5 , m = 4 ; int [][] arr={ { 1 , 2 , 3 , 4 }, { 5 , 6 , 7 , 8 }, { 9 , 10 , 11 , 12 }, { 13 , 14 , 15 , 16 }, { 17 , 18 , 19 , 20 }, }; // Function call diagonalOrder(arr, n, m); } } // This code is contributed by Manu Pathria |
Python3
R = 5 C = 5 def diagonalOrder(arr, n, m): # we will use a 2D vector to # store the diagonals of our array # the 2D vector will have (n+m-1) # rows that is equal to the number of # diagonals ans = [[] for i in range (n + m - 1 )] for i in range (m): for j in range (n): ans[i + j].append(arr[j][i]) for i in range ( len (ans)): for j in range ( len (ans[i])): print (ans[i][j], end = " " ) print () # Driver Code # we have a matrix of n rows # and m columns n = 5 m = 4 # Function call arr = [[ 1 , 2 , 3 , 4 ],[ 5 , 6 , 7 , 8 ],[ 9 , 10 , 11 , 12 ],[ 13 , 14 , 15 , 16 ],[ 17 , 18 , 19 , 20 ]] diagonalOrder(arr, n, m) # This code is contributed by rag2127 |
C#
using System; using System.Collections.Generic; public class GFG { public static int R = 5, C = 4; public static void diagonalOrder( int [,] arr, int n, int m) { // we will use a 2D vector to // store the diagonals of our array // the 2D vector will have (n+m-1) // rows that is equal to the number of // diagonals List<List< int >> ans = new List<List< int >>(n+m-1); for ( int i = 0; i < n + m - 1; i++) { ans.Add( new List< int >()); } for ( int i = 0; i < n; i++) { for ( int j = 0; j < m; j++) { (ans[i + j]).Add(arr[i, j]); } } for ( int i = 0; i < ans.Count; i++) { for ( int j = ans[i].Count - 1; j >= 0; j--) { Console.Write(ans[i][j] + " " ); } Console.WriteLine(); } } // Driver code static public void Main () { int n = 5, m = 4; int [,] arr={ { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 }, { 17, 18, 19, 20 }, }; // Function call diagonalOrder(arr, n, m); } } // This code is contributed by avanitrachhadiya2155 |
Javascript
<script> var R = 5; var C = 4; function diagonalOrder(arr, n, m) { // we will use a 2D vector to // store the diagonals of our array // the 2D vector will have (n+m-1) // rows that is equal to the number of // diagonals var ans = Array.from(Array(n+m-1), ()=>Array()); for ( var i = 0; i < n; i++) { for ( var j = 0; j < m; j++) { ans[i + j].push(arr[i][j]); } } for ( var i = 0; i < ans.length; i++) { for ( var j = ans[i].length - 1; j >= 0; j--) { document.write(ans[i][j] + " " ); } document.write( "<br>" ); } } // Driver Code // we have a matrix of n rows // and m columns var n = 5, m = 4; var arr = [ [ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ], [ 9, 10, 11, 12 ], [ 13, 14, 15, 16 ], [ 17, 18, 19, 20 ], ]; // Function call diagonalOrder(arr, n, m); // This code is contributed by rrrtnx. </script> |
输出
1 5 2 9 6 3 13 10 7 4 17 14 11 8 18 15 12 19 16 20
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