JAVA Best way to generate a line of " * "

Question

I am currently trying to generate a line of 60 stars however I found several ways to do it.

//Method 1
System.out.println(String.format("%60s", "").replace(' ', '*'));

//Method 2
for(int i=0;i<60;i++)
    System.out.print("*");
System.out.println("");

//Method 3
int i=0;
while(i<60) {
    System.out.print("*");
    i++;
}

Which is the best way to approach? (In terms of time and In terms of style)

And if there are any other methods to do this?

//Modular Approach, not bad
static String repeat(final String str, final int n) {
    final int len = (str == null) ? 0 : str.length();
    if (len < 1 || n < 1) {
        return "";
    }
    final StringBuilder sb = new StringBuilder(len * n);
    for (int i = 0; i < n; i++) {
        sb.append(str);
    }
    return sb.toString();
}

System.out.println(repeat("*", 60));

And

//Similar to method 1
System.out.println(new String(new char[60]).replace("\0", "*"));

Have you tried using `System.nanoTime()` to check how long each method takes? — Arc676, Nov 13 '15 at 03:45

Bohemian · Accepted Answer · 2015-11-13T04:15:58.110

How about:

Stream.generate(() -> "*").limit(60).forEach(System.out::print);
System.out.println(); // for the newline at the end

Or the slightly hacky, but one line:

System.out.println(new String(new char[60]).replace("\0", "*"));

The hacky version works because all java arrays of numeric primitives are initialized with the value 0 in all elements, which are each then replaced with the desired character once in a String.

score 2 · Answer 2 · answered Nov 13 '15 at 04:00

Of the methods presented, the first will likely perform best because it doesn't require a loop and only calls print once. But this smells of premature optimization. The choice of loop (for or while or do-while) is almost certainly optimized away by the compiler and JIT (depending on platform and version). However, I would prefer a fourth approach. Write Dumb Code. In this case, I would build the String with a StringBuilder and then print that. Something like,

static String repeat(final String str, final int n) {
    final int len = (str == null) ? 0 : str.length();
    if (len < 1 || n < 1) {
        return "";
    }
    final StringBuilder sb = new StringBuilder(len * n);
    for (int i = 0; i < n; i++) {
        sb.append(str);
    }
    return sb.toString();
}

and then

System.out.println(repeat("*", 60));

score 2 · Answer 3 · answered Nov 13 '15 at 04:13

2

Most efficient? How about:

int n=60;
char[] data = new char[60];

for(int i=0; i<n; i++)
    data[i] = '*';

System.out.print(data);

Only one object generated and a single call to "print".

answered Nov 13 '15 at 04:13

nimrodm

23,081
7
58
59

Agreed! How about `while(--n>=0)` instead of `for`? – T D Nguyen Nov 13 '15 at 04:20
It seems more complicated than method 2 – Bsonjin Nov 13 '15 at 05:11

score 2 · Answer 4 · answered Nov 13 '15 at 04:16

2

Unless you need the ability to easily modify the number of stars, I'd honestly just use System.out.println("************************************************************");

Or if you need to use the stars multiple times, which I'd assume is likely,

String sixtyStars = "************************************************************";

and then you can just pass sixtyStars to whatever printing method you need.

answered Nov 13 '15 at 04:16

z-ch

41
4

1

Haha nice, but i would prefer the stars which can be easily modified :) – Bsonjin Nov 13 '15 at 05:09

score 1 · Answer 5 · answered Nov 13 '15 at 04:05

In my opinions. . I think M1 is better and faster than using loops. However, M2 is easiest way to understand. Also you can add M4 by using

 PrintStart (60);\\ on call 


  \\function body 
 public static void PrintStart  ( int i){

       if (i==0)
           return; 
      System.out.print ("*");
      PrintStart  (i-1);
    }

T D Nguyen · Answer 6 · 2015-11-13T04:12:15.443

0

In terms of time complexity, they are the same: O(n). However M2=M3 better than M1 (It actually contains loops - you must see the source code of String.format and String.replace).

In terms of space complexity, M.2 is better.

In this problem M.2 is enough, can be improved by using StringBuilder.

edited Nov 13 '15 at 04:12

answered Nov 13 '15 at 03:57

T D Nguyen

7,054
4
51
71

ytterrr · Answer 7 · 2015-11-13T04:34:52.047

0

What about using custom collector?

String result = IntStream.range(0, 60).mapToObj(i -> '*')
        .collect(Collector.of(
                StringBuilder::new,
                StringBuilder::append,
                StringBuilder::append,
                StringBuilder::toString));

or the same with joining Strings

String result = IntStream.range(0, 60).mapToObj(i -> "*")
        .collect(Collectors.joining());

edited Nov 13 '15 at 04:34

answered Nov 13 '15 at 04:28

ytterrr

3,036
6
23
32

Kinda complicated but Thanks – Bsonjin Nov 13 '15 at 05:19

score 0 · Answer 8 · answered Nov 13 '15 at 05:16

0

Here's one I have used regularly - using Java 8 streams:

Stream.generate(() -> "*").limit(60).collect(Collectors.joining());

For me it captures the intent well: generate a stream of asterisks, get the first 60 and join them together.

answered Nov 13 '15 at 05:16

sprinter

27,148
6
47
78

Does this only work in Java 8 or above? – Bsonjin Nov 13 '15 at 05:19

JAVA Best way to generate a line of " * "

8 Answers8