Removing redundancy from String

[kmath]

Hello all,

I'm trying to eliminate the redundancy from this string using charAt. But can't quite nail it right. I'm trying to start with the first number in the string and cycle through all the others to make sure it's unique, and then if it is, write it out. Can you please take look and comment?

import java.io.*;
class charAt{
    public static void main(String[] args) {
String tf = "111111123422256";        
 
for (int i=0;i<(tf.length()-1);i++){
 
for (int z=0;z<(tf.length()-i);z++){
 
char c = tf.charAt(i); char k = tf.charAt(i+z);
 
		if(c==k){}
 
		else{
 
 
System.out.println(c);}}System.out.println(tf.charAt(tf.length()-1));
 
}}}

I realize this is off. The output should look like: 123456.

Please help. Thanks.

[angstrem]

What happens if i + z > tf.length?

[kmath]

fourteen 6s

[angstrem]

Why not to follow the following algorithm:
1. For a given character, check all the substring before this character
2. If the substring does not contain this character - print it
3. Repeat (1) - (2) for each character of the string

[pbrockway2]

I agree with angstrem. And, imho, some such "recipe" (algorithm) is the place to start rather than code.

[kmath]

I'm trying this now but still no luck:

import java.io.*;
class charAt{
    public static void main(String[] args) {
String tf = "111111123422256";        
 
for (int i=0;i<(tf.length());i++){
 
for (int k=0;k<i;k++){
 
if(tf.charAt(i)==tf.charAt(i-k)){}
 
		else{
 
System.out.println(tf.charAt(i));
 
 
}}}}}

[angstrem]

Say, you have string "12345". Say, i = 4 (that is, tf.charAt(i) == '5').
Now look at the code that checks all the substring:

for (int k=0;k<i;k++)
  if(tf.charAt(i)!=tf.charAt(k)) System.out.println(tf.charAt(i));

I've simplified it a bit to make it more readable, but this is essentially the same thing.
We have a loop. It checks each number before current. If it is distinct, it prints current number. In our case, all 4 numbers are distinct from '5'. Hence, '5' will be printed 4 times. Not something you want.