121 replies to this topic

[BenzR]

@BenzR: Please create a new thread in the Java Help section to get more help on this. You're hijacking a thread!

where is the help section 

Edited by dargueta, 28 April 2013 - 08:41 PM.

[fazilaayisha]

please help me solve this using nested for loops
15 14 13 12 11
10 9 8 7
6 5 4
3 2
1

[wim DC]

int number = 15;

for(int i=0 ; i<5 ; i++){

  for(int j=0 ; j<5-i ; j++){

    System.out.print(number--);

  }

}

[Majestic]

i am a rookie from java...i find it is quite hard for me:crying:
i wanna know how to print out like this?

88888881
88888812
88888123
88881234
88812345
88123456
81234567
12345678

could someone tell me the solution?

{
		int n = 8; int i; int j; 
		for (i=1;i<=n;i++) 
		{ 
			for (j=n;j>i;j--) 
				{ 
				System.out.print(n); 
				}
		
			for( j=1; j<=i;j++) 
			 { 
				System.out.print(j); 
			 }
		
			System.out.println();
		}
	} 

[fyhring4]

Thx this is really awesome!

[jun71178]

for you asma........ here is your request...

public class asma
{
    public static void main(String [] args)
    {
        for (int i=1; i<=4; i++)
        {
            
            for (int j=4; j>=i; j--)
            {
                System.out.print(i);
            }
        
      System.out.println(j);
    }
}

Edited by Alexander, 26 October 2010 - 09:49 AM.
Added [code] tags

[dandi]

City Map Block Builder

* if user input less than 1, will be defined as zero (0)

* user input the matrices

* Image of block :

  +---+
 /   /|
+---+ |
|   | +
|   |/
+---+

* Sample Input:

row : 3
column : 4

map matrices :
1 2 1 5
1 4 2 1
2 1 1 1

Result :

..................+---+
................./   /|
................+---+ |
................|   | +
................|   |/|
........+---+...+---+ |
......./   /|...|   | +
......+---+ |...|   |/|
......|   | +...+---+ |
......|   |/|-+.|   | +
......+---+ |/|.|   |/|
......|   | +---+---+ |
......|   |/   /|   | +
..+---+---+---+ |   |/|
./   /|   |   | +---+ |
+---+ |   |   |/   /| +
|   | +---+---+---+ |/.
|   |/   /   /   /| +..
+---+---+---+---+ |/...
|   |   |   |   | +....
|   |   |   |   |/.....
+---+---+---+---+......

how to make this

[siva]

how to make

1234
-234
--34
---4

[wim DC]

Solution for every loop your teacher is bothering you with.

It always consists of 1 outer for-loop with 1 or more inner for-loops.
The outer for-loop ALWAYS is the amount of rows you need. The inner loop will use System.out.print(); to print values.

So, let's take this sample:

88888881 
88888812 
88888123
88881234
88812345
88123456
81234567
12345678

The first things you want to do is count the number of rows: 8
Then you write this:

for( int i=0 ; i<[B][SIZE="4"]8[/SIZE][/B] ; i++){
  System.out.println("");
}

the println() will just make an "enter" / a newline.
The next thing you want to do is take the output and split the "equal stuff", okay badly explained but i hope the example helps:

8888888
888888
88888
8888
888
88
8

AND

1
12
123
1234
12345
123456
1234567
12345678

There are 2 things there. We start with the eights:
Now ask yourself how often it must print a value. You must find the link between the outer loop and the amount of times it must be printed.
Counter of the outer loop: i

i   #times to be printed
0         7
1         6
2         5
3         4
4         3
5         2
6         1
7         0

I hope everyone sees that #times to be printed is 7-i. If you don't see this, I suggest you go get some extra math classes.
7-i is important. because that's the link between the inner loop and outer loop.
The default for-loop is: for(int x=0 ; x<condition ; x++){}. Where x is a letter that's not used yet.

The condition to use in the default for-loop is 7-i as we've just found.
So put 7-1 as the condition (and I take the letter 'j' instead of 'x'):
This results is for(int j=0 ; j<7-i; j++){}
Because the value that is printed is constant, you are now able to add the 2nd loop to the code:

for( int i=0 ; i<8 ; i++){
  for( int j=0 ; j<7-i ; j++ ) {
     System.out.print("8");
  }
  System.out.println("");
}

Note it's a .print() in there, and no .printLN() we don't want newlines after every 8.

Now we need

1
12
123
1234
12345
123456
1234567
12345678

Counter of the outer loop: i

i   #times to be printed
0           1
1           2
2           3
3           4
4           5
5           6
6           7
7           8

IF you don't see here that #times to be printed = i+1, you really really need to take those math classes.
So we take our default for-loop again: for(int x=0 ; x<condition ; x++){}
condition = i+1
So fill in the condition (and I took letter 'k' instead of 'x' now)

for(int k=0 ; k<i+1; k++){
  System.out.print("?");
}

Unlike the previous loop the value that's printed is not constant. We need to figure out what it depends on.

Counter of the inner loop: k

i	k   #value to be printed
0	0           1
0	1           2
0	2           3
0	3           4
0	4           5
0	5           6
0	6           7
0	7           8
1       0           1
1       1           2
...

You could complete the table there but i hope it's allready clear.
Here we see we must print 'k+1', again if you fail to see this -->you really really need to take those math classes.
Now we can modify our last loop:

for(int k=0 ; k<i+1; k++){
  System.out.print([B]k+1[/B]);
}

And put it with the rest of the code in 1 piece:

    for( int i=0 ; i<8 ; i++){
      for( int j=0 ; j<7-i ; j++ ) {
        System.out.print("8");
      }
      for(int k=0 ; k<i+1; k++){
        System.out.print(k+1); 
      }
      System.out.println("");
    }

OUTPUT
------
88888881
88888812
88888123
88881234
88812345
88123456
81234567
12345678

Now let's take another example

15 14 13 12 11
10 9 8 7 
6 5 4
3 2 
1

Amount of rows:5

for(int i=0 ; i<5 ; i++){
  System.out.println("");
}

First look at the amount of times a value has to be printed:

i   #times to be printed
0           5
1           4
2           3
3           2
4           1

--> "#times to be printed = 5-i" if you fail to see this ... you know..
for(int x=0 ; x<condition ; x++){}
"5-i" = condition
result:

for(int j=0 ; j<5-i ; j++){
  System.out.print("?");
}

Now the value that needs to be printed
Counter of the outer loop: i
Counter of the inner loop: k

i    k   #value to be printed
0    0           15                
0    1           14               
0    2           13            
0    3           12                         
0    4           11
---next loop---
1    0           10
1    1           9
1    2           8
1    3           7
---next loop---
2    0           6
2    1           5
2    2           4
---next loop---
3    0           3
3    1           2
---next loop---
4    0           1

Here the value that has to be printed has nothing to do with anything. It's just a number 15 that's lowered by 1 every loop.
So create a number 15 OUTSIDE all loops.
Print and lower this number every loop

int number = 15;
-----above all loops-----

-----inside loop-----
for(int j=0 ; j<5-i ; j++){
  System.out.print(number + " ");
  number = number -1;
}

(don't forget to print the space)
Final result:

int number=15;
    for(int i=0 ; i<5 ; i++){
      for(int j=0 ; j<5-i ; j++){
        System.out.print(number +" ");
        number = number -1;
      }
      System.out.println("");
    }

OUTPUT
------
15 14 13 12 11 
10 9 8 7 
6 5 4 
3 2 
1 

Are we done yet? ** NO! another short example:

1
22
333
4444
55555

Amount of rows: 5

for(int i=0 ; i<5 ; i++){
  System.out.println("");
}

How many times is it printed each row?

0    1
1    2
2    3
3    4
4    5

So this is i+1 = condition
fill in condition in for(int x=0 ; x<condition ; x++){}

for(int j=0 ; j<i+1 ; j++){
  System.out.print("?");
}

What values are printed each row?

0    1
1    2
2    3
3    4
4    5

so this is also i+1
adapt loop:

for(int j=0 ; j<i+1 ; j++){
  System.out.print(i+1);
}

Put into outer loop:

for(int i=0 ; i<5 ; i++){
  for(int j=0 ; j<i+1 ; j++){
    System.out.print(i+1);
  }
  System.out.println("");
}

TADAAA, done.

Recap
most of the times:

• Amount of rows -> outer loop
• How many times per loop? --> inner loop condition
• What has to be printed --> in the System.out.print(<--->);

Now I don't wanna see requests here with no code. At least the outer loop is EASY.

Edited by wim DC, 29 November 2010 - 09:01 AM.

[siva]

I need help on how to write a method that will return the reverse of an inputted word
public String reverse (String word)
{

}

[wim DC]

I need help on how to write a method that will return the reverse of an inputted word
public String reverse (String word)
{

}

That's very simple because the StringBuilder class has a .reverse() method.

StringBuilder sb = new StringBuilder("MyWord");
String reverse = sb.reverse().toString();
System.out.println(reverse);


---output---
droWyM

Didn't really test this code but i suppose it's correct.

[Alexander]

Nice compilation oxano, I was like "Woah", wonder if this should be a new version of this outdated tutorial