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How to separate a string into substrings of two characters

I am trying to split strings in substrings of two chararters for example for the input: "ABCDE" i want to get the substrings "AB" "BC" "CD" "DE".
I tried with this:
String route = "ABCDE";
int i = 0;
while(i < route.length()) {
String sub = route.substring(i,i+2);
System.out.println(sub);
i++;
}
but the index (i) gets out of range int the last iteration and causes an error.
is there any way to do this without getting the index (i) out of range ?

You need to change the loop condition.
while(i < route.length()-1)
In your code i goes till (length-1) and than in the substring(i,i+2) function you gives end index i+2. It is higher than largest index of string.
Also, As far as I know calling a library function in a loop condition is not considered a good practice.
In each iteration you call this function which is time consuming.
control goes to that subroutine in each iteration.
A good alternative to this would be to store the length in a variable and use that in a condition.
int temp = route.length()-1;
while(i<temp){

This should work fine
String route = "ABCDE";
if( route.length() > 2){
int i = 0;
do {
String res = route.substring(i,i+2);
System.out.println(res);
i++;
} while (i + 1 < route.length());
}
else{
System.out.println(route);
}
Edit: Added boundary case for the string has length less than 2

Add check for the size of the string to trap the error:
String route = "ABCDE";
int i = 0;
while(i < route.length()) {
if(i < route.length() - 1) {
String sub = route.substring(i,i+2);
System.out.println(sub);
} else {
String sub = route.substring(i,i+1);
System.out.println(sub);
i++;
}
So whenever the i counter almost close to string size, get the last char.

You are getting an StringIndexOutOfBoundsException because you are trying to access an index of the String that doesn't exist.
To fix this, change your loop condition from
while(i < route.length())
to
while(i < route.length() - 1)
Without the -1 on the last iteration of the while loop i + 2 is equal to 71 which is out of the Strings bounds.
Another (cleaner) solution to this problem is a for loop:
for(int i = 0; i < route.length() - 1; i++) {
System.out.println(route.substring(j, j + 2));
}
The for loop in this situation is just shorter as the declaration, conditional, and increment statements are all in one line.
1: This 7 reduces to 6 since the endIndex of substring is exclusive.

As denis already pointed out, the bug in the code is in the loop condition.
Should be: while(i < route.length() - 1)
. However, how about simplifying this logic to use a for loop.
String route = "ABCDE";
for (int i=0; i+2<=route.length(); i++)
System.out.println(route.substring(i,i+2));

you can not use
i < route.length(),because when i = 5, String sub = route.substring(i,i+2); the i+2=7,is out of index,so use i<route.length instead

How can I reprint a string starting from the first character till the last?

Given a non-empty string str like "Code" print a string like "CCoCodCode". Where at each index in the string you have to reprint the string up to that index.
I know there is DEFINITELY something wrong with this code that I wrote because the answer should be CCoCodCode, but instead it's giving me the alphabet! I don't know how I should change it.
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
String str = scan.next();
int x = str.length();
for(char i = str.charAt(0); i <= str.charAt(x-1); i++)
{
System.out.print(i);
}
}

The char datatype can be treated as a number; you can increment it and manipulate it as a number.
What you really want is successive substrings of str to be printed. Loop over an int that will represent the ending position of the substring to be printed.
for (int i = 0; i < str.length(); i++)
{
System.out.print(str.substring(0, i + 1));
}
The end index argument to substring is exclusive, which is why I added 1.

Let's say that str is "Code". We can perform some mental substitutions to see what happens to your loop.
str is "Code"
x is 4
str.charAt(0) is 'C'
str.charAt(x-1) is 'e'
Making these substitutions, your loop is:
for(char i = 'C'; i <= 'e'; i++)
{
System.out.print(i);
}
Does this help you see the problem? I would think you'd have a loop from 0 to 3, not from 'C' to 'e'...

Many ways to get it done, suppose we have the input from user stored in a string named "c"... then...
String c = "Code";
for (int i = 0; i < c.length(); i++) {
System.out.print(c.substring(0, i));
}
System.out.print(c);
And this will print the sequence you are looking for.

It is outputting the alphabet because you are printing the counter instead of the characters in the string!
As it is, the first iteration of the for loop will set i to the first character, print that, then the operation i++ will increment i by one. Wait, so if the first character is "C", so i = 'C', what is i++?
Well it turns out characters can be represented by numbers. For example, 'C' has a value of 67. So incrementing it makes it 68, which represents 'D'. So if you run the loop on "Code", it will increment your counter 4 times, giving "CDEF". If you run on "Codecodecode", that will make the loop run 12 times, giving "CDEFGHIJKLMN".
What you really want is to loop through the string by its index instead:
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String str = scan.next();
int length = str.length();
for (int i = 0; i < length; i++) {
System.out.print(str.substring(0, i + 1));
}
}

Palindromes using Stacks and Queues

My NEW sample text i was testing: My mom is a good cook. Although sometimes at around noon she will leave and forget to make me lunch and some pop. #Old homework become relevant again
my problem is just that i am not getting the correct output, as my method only prints *Mom a noon i
This is all GUI based.I am reading in a file and checking for palindromes and printing them out in my JTextArea afterwards using Stacks and Queue's.
Issue is, all of this is working and when i start the program and attach the text file, i only get the first palindrome. SO it will print "mom" which is my first testcase, but it won't go to any of the other palindromes following it?
I thought i might have got bogged down in my code blocking at some point but after tinkering with it for a day now i'm sort of stuck.
EDIT 1: I am now getting more results
my method is,
public void fileDecode() throws FileNotFoundException
{
if (fileChoice.showOpenDialog(null) == JFileChooser.APPROVE_OPTION)
{
file = fileChoice.getSelectedFile();
scanInput = new Scanner(file);
while(scanInput.hasNext())
{
int nextPalindrome = 0;
int counter = 0;
String token = scanInput.next();
Stack<Character> stk = new Stack<Character>();
Queue<Character> que = new LinkedList<Character>();
for (int i = 0; i < token.length(); ++i)
{
stk.push(token.charAt(i)); //pushing all char's onto the stack/queue
que.add(token.charAt(i));
}
for (int j = 0; j < token.length(); ++j)
{
char tempStk = stk.pop(); //removing them one at a time and checking if they are equal and incrementing counter
char tempQue = que.remove();
if (tempStk == tempQue)
{
counter++;
}
}
if (counter == token.length())
{
build.append(token + " "); //if the counter # was the same as the token's length, than it is indeed a palindrome and we append it into our StringBuilder that we pass to the JTextArea
nextPalindrome = token.length();
}
}
}
}

You set counter to 0 outside your while loop, so the count of the second word is going to start at the count of the first word. Move counter = 0 inside the while loop and it should work.
Also, nextPalindrome doesn't appear to be used, and even if it is, if you set it at the bottom of the loop, it's immediately set to 0 at the top, so it will only remain non-zero if the last word is a palindrome.
Also, think about what's happening in the second for loop. You're looping over all the characters and comparing the one from the stack and the one from the queue. If ever those are different, you know you don't have a palindrome, so once you find a difference, it's pointless to continue with the loop. You also already have a loop counter, j, so you don't need another one. So I'd rewrite the second loop and following condition as follows:
for (int j = 0; j < token.length(); ++j)
{
char tempStk = stk.pop(); //removing them one at a time and checking if they are equal and incrementing counter
char tempQue = que.remove();
if (tempStk != tempQue)
break;
}
if (j == token.length())
That works because the only way j can equal token.length() after the loop is done is if the loop completed, which can only happen if no pairs of characters aren't equal (in other words, all pairs of characters are equal, which is what you want).

Java infinite loop issue

So our teacher gave us this homework, we had to write a program that went something along the lines of
Write an application that reads a line of text from the keyboard and prints a table indicating the number of occurrences of each letter
of the alphabet in the text, For example, the phrase
To be, or not to be: that is the question:
Contains one “a,” two “b’s,” no “c’s,” and so on.
Well I've written the code, but I've ran into one small problem when I enter the to be or not to be part the code continually loops forever. I've looked at this program forever, I even tried asking some folks at Yahoo (but I think I confused them). So I am hoping someone here will spot something I missed or have some advice to give me.
public class occurances {
public static void main(String[] args) {
Scanner inp = new Scanner(System.in);
String str;
char ch;
int count = 0;
System.out.println("Enter the string:");
str = inp.nextLine();
while (str.length() > 0) {
ch = str.charAt(0);
int i = 0;
while (i < str.length() && str.charAt(i) == ch) {
count = count++;
i++;
}
str = str.substring(count);
System.out.println(ch);
System.out.println(count);
}
}
}

There are many problems with your code, but start with count = count++. It will always have its initial values (0 in your case). That causes the infinite loop. If you manage this one you'll be good to go with debugging your code further. Learn how to use debugger and/or print for checking your code. Good luck.

It seems your approach is not necessarily the one the teacher wants. Your approach (if it worked at all) would display the character counts in the order the characters appear in the string. So, for example, for "To be, or not to be: that is the question" you would show the character count for "T" first, whereas the teacher probably wants you to show the character count for "a" first. Your approach also doesn't show the character counts for the characters that are missing in the answer.
It has been suggested in the other two answers to use a Map. I recommend that approach, although you could use a simple int[] array where the index is (ch - 'a') assuming that it is between 0 and 25. See Character.toLowerCase() for how to convert a character into a lowercase one, because the correct answer would probably treat "T" and "t" as the same.
You need only one loop through the array, incrementing the counts for the characters that appear. Obviously all of the counts should be initialized to 0 prior to the loop.

I've not done JAVA in a while i came here to help someone with Javascript... but i'll give it go
while(str.length()>0)
This will obviously enter an endless loop because you have no breakout terms... i'd probably use a for loop instead... also you need something to store your counts inside of for each unique letter you find.
-- EDIT based on the comment below i find myself agreeing that answering your homework is not a good thing but here is what i'd suggest you look at...
Look at Hashtables or Maps and thing about "storing the dictionary counts as you find them"
Bare in mind upper and lower case letter, i'd just cast to lowercase and count
use a for loop instead of the while loop, use the strings length as the limit e.g. count from 0 to string.length()
This should get you on the right path :http://www.java2s.com/Tutorial/Java/0140__Collections/GettingelementskeyvaluepairsfromaHashtabletheentrySetmethod.htm

Some things to resolve in your code:
while(str.length()>0) : Infinite loop here. You could go through the string, with a for (int index = 0; index < str.length(); index++).
ch = str.charAt(0); : Here you are always taking the first letter of the String, you should take each letters with ch = str.charAt(index);.
count = count++; : count++; would be enough or count = count + 1;.
count = count++; : count should be re-initialized for each letter count = 0; before your while inner loop.
str = str.substring(count); : No need to substring here. You would loose letters for next computation.
Your code should work with one disavantages. The letters and their count will be written as many times as their is their occurence. If a is 3 times in the String, a : 3 will be written 3 times.
So you should whether you had already print the letter (maybe an array), but performance would be bad. For a String of n letters, you would have n * n computations (Complexity O(n^2)). If you take an HashMap between letter and their count, you will just of to go through the String one time, incrementing your counter while iterating on the loop. Then you would have only n computations. (Complexity O(n)).
HashMap<Char, Integer> letters = new HashMap<Char, Integer>();
for (int i = 0; i < str.length(); i++) {
char letter = str.charAt(i);
if (!letters.contains(letter)) {
letters.put(letter, 1);
} else {
letters.put(letter, letters.get(letter) + 1);
}
}
Then you go through the map to display the count.

Here is your corrected code for the problem you asked..Try this..
import java.util.*;
public class Occurence {
public static void main(String[] args) {
Scanner inp = new Scanner(System.in);
String str;
char ch;
int count = 0;
System.out.println("Enter the string:");
str = inp.nextLine();
int length = str.length();
int i = 0, j = 0;
for (i = 0; i < length; i++) {
j = 0;
count=0;
for (j = 0; j < length; j++) {
if (str.charAt(j) == str.charAt(i)) {
count++;
}
System.out.println(str.charAt(i)+" "+count);
}
}
}
}

Loop with incremental value based on user input

I have a task about incremental values in a loop based on user input.
The task is that the following lines are generated in the console
*
**
***
****
*****
And the amount of lines are decided by user input. I.e. if the user types in 2 it gives the following output:
*
**
My current code is:
import static javax.swing.JOptionPane.*;
public class loop1 {
public static void main(String[] args){
int current_line = 0;
String input_line = showInputDialog("type here ");
int target_line = Integer.parseInt(input_line);
while (current_line != target_line){
String sign = "*";
System.out.println(sign);
current_line ++;
}
}
}
But I can't seem to get the number of asterisks (*) to increase for every time it runs. How can I accomplish this?

You need a nested loop. Each iteration of the outer loop (which is the loop you already have) would print a single row, and each iteration of the inner loop would print a single asterisk for the current row.

You actually need two loops here, but you only have one. You have a while loop to print out the asterisks, but you also need a loop to increment the number of asterisks printed out each time.
Some pseudocode might look like:
For (int i = 1 to whatever value the user entered):
For (int j = 1 to i):
Print an asterisk
Actual code would look like:
int numLines = Integer.parseInt(showInputDialog("type here "));
for(int numAsterisks = 0; numAsterisks < numLines; numAsterisks++) {
for(int i = 0; i < numAsterisks; i++) {
System.out.print("*");
}
System.out.println(); // Start a new line
}

You can make it simpler by using nested for loops. Modify your loop to:
for (int i=0;i<target_line;i++) {
for (int j=0;j<=i;j++) {
System.out.print("*");
}
System.out.println();
}

You print everytime one '*'-sign.
You don't necessarily need two loops. You can place the sign outside of the loop and you can add an asterisk every iteration with string.concat("*"). Concatenating actually means combining two strings into one, so you actually combine the sign from the previous iteration with a new sign.
int current_line = 0;
String input_line = showInputDialog("type here ");
int target_line = Integer.parseInt(input_line);
String sign = "*";
while (current_line != target_line){
sign.concat("*");
System.out.println(sign);
current_line ++;
}