通过二维数组对角循环 [英] Loop diagonally through two dimensional array

问题描述

  String [] [] b = [a，b，c] 
 [d，e，f] 
 [g，h，i]; 
 $ b $ 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）; 
 
 
 
 
 
 
现在输出对角线即当前输出： / p> 
 
 
  g dg aei bf c 
  
如何使其打印其他对角线，即所需的输出：
 
 $ p $  a db gec hf i 
  




解决方案

初始化数组仅用于测试目的b
$ b

  int dim = 5; 
 char ch ='A'; 
 String [] [] array = new String [dim] []; 
 for（int i = 0; i  array [i] = new String [dim]; 
 for（int j = 0; j  array [i] [j] =+ ch; 
 
 
 
 
 
 
 
输出我们的矩阵：$ b $ （int j = 0; j  for（int j = 0; j  System.out.print（array [i] [j] +）; 
} 
 System.out.println（）; 
} 
 System.out.println（============================）; 
  





解决方案

对角线元素索引一个规则 - 它们的总和在一个对角线上是不变的：

VARIANT 1

使用两个循环提取所有对角线。

第一个循环提取对角线的上半部分：

  for int k = 0; k  for（int j = 0; j <= k; j ++）{
 int i = k-j; 
 System.out.print（array [i] [j] +）; 
} 
 System.out.println（）; 
 
 $ / code> 

第二个循环遍历对角线的下半部分：$ b $对于（int j = 0; j< b
pre $ ; = k; j ++）{
int i = k - j;
System.out.print（array [dim - j - 1] [dim - i - 1] +）;
}
System.out.println（）;



$ b $ h $ VARIANT 2



使用一个循环来提取所有的对角线，但是有额外的迭代和一个额外的检查：



int i），（int k = 0; k > = k - j; （i System.out.print（array [i] [j] +）;
}
}
System.out.println（）;

输出：

  ABCDE 
 FGHIJ 
 KLMNO 
 PQRST 
 UVWXY 
 ============== ============== 
 A 
 FB 
 KGC 
 PLHD 
 UQMIE 
 VRNJ 
 WSO 
 XT 
 Y 
  

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 dg aei bf c

How do I make it print the other diagonals i.e required output:

a db gec hf i 

解决方案

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 

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