问题描述
I wrote the following code to walk half the diagonals of an array:
String[][] b = [a,b,c]
[d,e,f]
[g,h,i];
public void LoopDiag()
for (int i = b.length - 1; i > 0; i--) {
String temp = "";
for (int j = 0, x = i; x <= b.length - 1; j++, x++) {
temp = temp+b[x][j];
}
System.out.println(temp)
}
for (int i = 0; i <= b.length - 1; i++) {
String temp = "";
for (int j = 0, y = i; y <= b.length - 1; j++, y++) {
temp = temp+b[j][y];
}
System.out.println(temp);
}
}
Right now it prints the diagonals i.e current output:
g dh aei bf c
How do I make it print the other half diagonals i.e required output:
a db gec hf i
Initialize array only for test purpose:
int dim = 5;
char ch = 'A';
String[][] array = new String[dim][];
for( int i = 0 ; i < dim ; i++ ) {
array[i] = new String[dim];
for( int j = 0 ; j < dim ; j++, ch++ ) {
array[i][j] = "" + ch;
}
}
Output our matrix:
for( int i = 0 ; i < dim ; i++ ) {
for( int j = 0 ; j < dim ; j++, ch++ ) {
System.out.print( array[i][j] + " " );
}
System.out.println();
}
System.out.println( "============================" );
Solution
Element indexes from diagonals have one rule - their sum is constant on one diagonal:
VARIANT 1
Use two loops to extract all diagonals.
First loop extracts top half of diagonals:
for( int k = 0 ; k < dim ; k++ ) {
for( int j = 0 ; j <= k ; j++ ) {
int i = k - j;
System.out.print( array[i][j] + " " );
}
System.out.println();
}
Second loop iterates on bottom half of diagonals:
for( int k = dim - 2 ; k >= 0 ; k-- ) {
for( int j = 0 ; j <= k ; j++ ) {
int i = k - j;
System.out.print( array[dim - j - 1][dim - i - 1] + " " );
}
System.out.println();
}
VARIANT 2
Use one loop to extract all diagonals, but there are extra iterations and one additional check:
for( int k = 0 ; k < dim * 2 ; k++ ) {
for( int j = 0 ; j <= k ; j++ ) {
int i = k - j;
if( i < dim && j < dim ) {
System.out.print( array[i][j] + " " );
}
}
System.out.println();
}
The output:
A B C D E
F G H I J
K L M N O
P Q R S T
U V W X Y
============================
A
F B
K G C
P L H D
U Q M I E
V R N J
W S O
X T
Y
Update
There are questions about rectangular matrix (height != width) in comments. Here is solution for rectangular matrix:
The rule remains the same: Sum of element indexes from the same diagonal is constant
The minimum sum of indexes is 0 (for first element in matrix with indexes [0;0])
The maximum sum of indexes is width + height - 2 (for last element in matrix with indexes [height-1; with-1])
Initialize rectangular matrix only for test purpose:
int WIDTH = 7;
int HEIGHT = 3;
char ch = 'A';
String[][] array = new String[HEIGHT][];
for( int i = 0 ; i < HEIGHT ; i++ ) {
array[i] = new String[WIDTH];
for( int j = 0 ; j < WIDTH ; j++, ch++ ) {
array[i][j] = "" + ch;
}
}
Print our rectangular matrix:
for( int i = 0 ; i < HEIGHT ; i++ ) {
for( int j = 0 ; j < WIDTH ; j++, ch++ ) {
System.out.print( array[i][j] + " " );
}
System.out.println();
}
System.out.println( "============================" );
Solution
for( int k = 0 ; k <= WIDTH + HEIGHT - 2; k++ ) {
for( int j = 0 ; j <= k ; j++ ) {
int i = k - j;
if( i < HEIGHT && j < WIDTH ) {
System.out.print( array[i][j] + " " );
}
}
System.out.println();
}
Output:
A B C D E F G
H I J K L M N
O P Q R S T U
============================
A
H B
O I C
P J D
Q K E
R L F
S M G
T N
U
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