for loop, iteration through alphabet? java

Question

I can iterate through the alphabet, but I'm trying to keep the last iteration and add on the next letter. this is my code.

for(char alphabet = 'a'; alphabet <='z'; alphabet ++ )
        {

            System.out.println(alphabet);
        }

I want it to print out something that looks like this.

a

ab

abc

abcd

abcde..... and so forth. How is this possible?

Solution 1

You need to add the char alphabet to a string.

String output = "";

for(char alphabet = 'a'; alphabet <='z'; alphabet++ )
    {
        output += alphabet;
        System.out.println(output);
    }

This should work for you ;)

Solution 2

I will go with StringBuffer or StringBuilder. Something like:

StringBuffer

StringBuffer sb = new StringBuffer();
for (char alphabet = 'a'; alphabet <= 'z'; alphabet++) {
    sb.append(alphabet);
    System.out.println(sb.toString());
}

StringBuilder

StringBuilder sb = new StringBuilder();
for (char alphabet = 'a'; alphabet <= 'z'; alphabet++) {
    sb.append(alphabet);
    System.out.println(sb.toString());
}

String vs StringBuffer vs StringBuilder:

String: It is immutable, so when you do any modification in the string, it will create new instance and will eatup memory too fast.

StringBuffer: You can use it to create dynamic String and at the sametime only 1 object will be there so very less memory will be used. It is synchronized (which makes it slower).

StringBuilder: It is similar to StringBuffer. The olny difference is: it not synchronized and hence faster.

So, better choice would be StringBuilder. Read more.

Using Java 8

StringBuilder sb = new StringBuilder();

IntStream.range('a', 'z').forEach(i -> {
    sb.append((char) i);
    System.out.println(sb.toString());
});

Solution 3

I'd suggest using a StringBuilder:

// Using StringBuilder
StringBuilder buf = new StringBuilder();
for (char c = 'a'; c <= 'z'; c++)
    System.out.println(buf.append(c).toString());

You could also do it slightly faster by using a char[], however StringBuilder is more obvious and easier to use:

// Using char[]
char[] arr = new char[26];
for (int i = 0; i < 26; i++) {
    arr[i] = (char)('a' + i);
    System.out.println(new String(arr, 0, i + 1));
}

Alternatives that you shouldn't use:

• StringBuffer: Same as StringBuilder, but synchronized, so slower.
• s = s.concat(new String(c)): Allocates 2 Strings per iteration, instead of only 1.
• s += c: Internally += is compiled to s = new StringBuilder().append(s).append(c).toString(), so horrendously slow with exponential response times.

Comments

[Andreas]

Never use String += String in a loop. Use StringBuilder. Sure, this is a small inconsequential loop, but start doing it the right way, so it becomes second nature.

[pewpew]

i edited like 2 seconds after i posted lol! such fast response time haha. One thing to note. the output += alphabet needs to be inside the print statement System.out.println(output += alphabet);

[Andreas]

StringBuffer is obsolete. Use StringBuilder. Quoting javadoc: "As of release JDK 5, StringBuffer has been supplemented with an equivalent class designed for use by a single thread, StringBuilder."

[Stefan]

I don't think that has to be in the System.out. Why do you think so?

[Ambrish]

I have added both in my solution.

[Andreas]

@pewpew Sure that's shorter, but "needs to be"? No.

[Rahman]

I am assuming he will declare it outside of that for loop :). Anyways edited the answer

[Andreas]

StringBuffer is synchronized, which makes it slower. Unless you're in an extremely rare situation where multiple threads append to the same buffer, you should never use it. Always use StringBuilder. Make it a habit. I'd suggest removing StringBuffer from answer, so as not to misguide any newbie Java developers coming here.

[singhakash]

@Andreas I dont suggest removing I would suggest the Ambrish should clarify the difference so that OP can choose any method accoding the requirement

[pewpew]

Ok i see what you did there ;)

[Ambrish]

@Andreas I have added details and also better solution: Java8.

[Andreas]

Java 8 stream is a nice touch, but I disagree with calling it a "better solution". To me, it's actually slightly worse, but I'd be good with simply calling it "another" way of doing it.

[Andreas]

@pewpew Of all the answers, you accept the one with the worst performance of them all. It is a bit simpler to write than the rest, I'll give you that. It's a good thing there's only 26 letters in the alphabet.

[Vogel612]

While this may theoretically answer the question, it's not really a good answer, since it doesn't teach the OP. Instead it gives an alternative solution without explanation. This will usually lead to OP not learning, and coming back for asking a new question when a similar problem occurs. Would you mind adding some explanation?