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Guessing symbols in a Sting word java [closed]

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Got a String like:
String str = "###############";
Got guess word, for example:
String guess = "Java"
User must guess word:
User input:
Sava
Sring should be:
String str = "#a#a###########";
all right symbols placed on their indexes
String is immutable class.
I chose Stringbuilder
for (int i = 0; i < length ; i++) {
if (rnd.charAt(i) == guess.charAt(i) && rnd.charAt(i) != '#'){
sb.append(rnd.charAt(i));
}
}
System.out.println(sb);
sb.delete(0, sb.length());
Stringbuilder add right symbols not on possition 'i', but on the last indexes.
Example:
guess word: Java
user input Sala:
System.out.println(sb);
###############aa
How I can achieve needed result?
And what tools should I use?
needed result:
Example:
guess word Java:
user input Sala:
System.out.println(sb);
#a#a###########

Work like this:
private static String word(){
String guess = new Scanner(System.in).nextLine();
return guess;
}
private static void guessWord(String[]arr) {
int random = new Random().nextInt(arr.length);
String rnd = arr[random];
int length = 15;
StringBuilder sb = new StringBuilder();
String guess = "";
int rndLength = length - rnd.length();
int guessLength = length - guess.length();
do {
System.out.println("Enter a word: ");
guess = word();
if (sb.length() < length){
for (int i = 0; i < length ; i++) {
sb.append("#");
}
}
for (int i = 0; i < length && i < rnd.length() && i < guess.length(); i++) {
if (rnd.charAt(i) == guess.charAt(i)){
sb.setCharAt(i, rnd.charAt(i));
sb.delete(length, sb.length());
}
}
if (rnd.equals(guess)){
System.out.println("Guess word: " + rnd);
break;
}else if (!rnd.equals(guess)) {
System.out.println(sb);
}
}while (!rnd.equals(guess));
}

You can do it as follows:
public class Main {
public static void main(String[] args) {
String str = "#a#a###########";
String guess = "Java";
String input = "Sala";
StringBuilder sb = new StringBuilder();
int i;
for (i = 0; i < str.length() && i < guess.length() && i < input.length(); i++) {
// In case of a match, append the matched character
if (guess.charAt(i) == input.charAt(i)) {
sb.append(guess.charAt(i));
} else {// Else append the placeholder symbol from `str`
sb.append(str.charAt(i));
}
}
// Append the remaining placeholder characters from `str`
sb.append(str.substring(i));
// Display
System.out.println(sb);
}
}
Output:
#a#a###########

How can I compare these two strings and remove a common letter without using arrays?

This was the code I designed to solve this problem but it seems not to work at all.I used nested for loops to compare the letters of the first string and the second string since they are likely to have different lengths
import java.util.*;
public class Trim
{
public static String myTrim(String input, String list)
{
String r = "";
for (int i = 1; i < input.length();i++)
{
for (int k = 1; k < list.length();k++)
{
if (input.charAt(i) != list.charAt(i))
{
r += input.charAt(i);
}
}
}
return r;
}
}

I guess you should use the method String.indexOf.
So:
public static String myTrim(String input, String list)
{
StringBuilder result = new StringBuilder();
char c;
for (int i = 0; i < input.length(); i++)
{
c = input.charAt(i);
if (list.indexOf(c) < 0)
result.append(c);
}
return result.toString();
}

Try using a flag to determine whether to character gets repeated or not:
String r = "";
for (int i = input.length() - 1; 0 <= i; i --) {
if (-1 == list.indexOf(input.charAt(i))) {
r += input.charAt(i);
}
}
OR
String r = "";
boolean found;
for (int i = input.length() - 1, j = list.length() - 1; 0 <= i; i--) {
found = false;
for (int k = j; 0 <= k; k--) {
if (list.charAt(k) == input.charAt(i)) {
found = true;
break;
}
}
if (!found) {
r += input.charAt(i);
}
}

We have to filter out the characters from input which appears in list.
Now we have to check whether each character of the input appears in the list or not.
The k value will be less then list.length() if the character of input present in the list string.
After the loop we check the k value and append it to the new string.
public static String myTrim(String input, String list)
{
String r = "";
for (int i = 0; i < input.length();i++)
{
int k = 0;
for (; k < list.length();k++)
{
if (input.charAt(i) == list.charAt(k))
{
break;
}
}
if(k == list.length())
r += input.charAt(i);
}
return r;
}

A nice one-liner solution would be to use Guava Charmatcher:
CharMatcher.anyOf(list).removeFrom(input);

I have tried this code and it's working fine with both of your inputs
for (int i = 0; i < S1.length(); i++) {
for(int j=0;j< S2.length();j++) {
if(S1.charAt(i)==S2.charAt(j)) {
char Temp= S2.charAt(j);
String Temp2=String.valueOf(Temp);
S1=S1.replace(Temp2, "");
}
}
}
This is code
import java.util.Scanner;
public class StringRemoveChar {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String S1, S2;
S1 = scanner.nextLine();
S2 = scanner.nextLine();
for (int i = 0; i < S1.length(); i++) {
for (int j = 0; j < S2.length(); j++) {
if (S1.charAt(i) == S2.charAt(j)) {
char Temp = S2.charAt(j);
String Temp2 = String.valueOf(Temp);
S1 = S1.replace(Temp2, "");
System.out.println(S1.length());
}
}
}
System.out.println(S1);
}
}
Input:
Miyazaki
you
Output:
Miazaki

We can use replaceAll and use one loop over ,this will make the solution simple
public static String myTrim(String input, String list)
{
for(int i=0;i<list.length();i++)
{
input=input.replaceAll(list.charAt(i)+"","");
}
return input;
}
Input: myTrim("Miyazaki","you")
Output: Miazaki
Full code for reference
package stackoverflow.string;
public class StringManipulation
{
public static void main(String[] args)
{
System.out.println(myTrim("University of Miami","city"));
}
public static String myTrim(String input, String list)
{
for(int i=0;i<list.length();i++)
{
input=input.replaceAll(list.charAt(i)+"","");
}
return input;
}
}

Check if a string has all the characters of another string [closed]

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String one = "This is a test";
String two = "This is a simple test";
I want to check if two contains all the characters that are in one, and ignore the fact it has extra characters.

The fastest thing would probably be to break them up to HashSets and then apply containsAll
public static Set<Character> stringToCharacterSet(String s) {
Set<Character> set = new HashSet<>();
for (char c : s.toCharArray()) {
set.add(c);
}
return set;
}
public static boolean containsAllChars
(String container, String containee) {
return stringToCharacterSet(container).containsAll
(stringToCharacterSet(containee));
}
public static void main(String[] args) {
String one = "This is a test";
String two = "This is a simple test";
System.out.println (containsAllChars(one, two));
}

static boolean stringContains(String longer, String shorter) {
int i = 0;
for (char c : shorter.toCharArray()) {
i = longer.indexOf(c, i) + 1;
if (i <= 0) { return false; }
}
return true;
}

Using a simple loop over the set of the characters in the first string:
String s1 = "This is a test";
String s2 = "This is a simple test";
Set<Character> chars = new HashSet<Character>();
for(int i = 0; i < s1.length(); i++) {
chars.add(s1.charAt(i));
}
for (Iterator<Character> iterator = chars.iterator(); iterator.hasNext();) {
Character character = iterator.next();
if(!s2.contains(character.toString())) {
// break and mark as not contained
break;
}
}
If words are meant to be checked, then you could split the string around whitespaces into a list of words:
String[] words1 = s1.split("\\s");
String[] words2 = s2.split("\\s");
List<String> wordList1 = Arrays.asList(words1);
List<String> wordList2 = Arrays.asList(words2);
System.out.println(wordList2.containsAll(wordList1));

two.startsWith(one)
If you aren't sure about the start position (the above assumes 0), try using the following API
startsWith(String, offset)

Try this. I know it's long but it works
public static void main(String[] args)
{
String String1,String2;
int i, j, count1, count2;
count1 = 0;
count2 = 0;
String1 = "This is a test";
String2 = "This is a simple test";
char[] list1 = new char[String2.length()];
char[] list2 = new char[String2.length()];
char[] list3 = new char[String2.length()];
for (i = 0; i <= String1.length() - 1; i++)
{
list1[i] = String1.charAt(i);
for (j = 0; j <= String2.length() - 1; j++)
{
list2[j] = String2.charAt(j);
if (list1[i] == list2[j])
{
i++;
count1++;
}
}
}
for (i = 0; i <= String1.length() - 1; i++)
{
list1[i] = String1.charAt(i);
for (j = 0; j <= String1.length() - 1; j++)
{
list3[j] = String1.charAt(j);
if (list1[i] == list3[j])
{
i++;
count2++;
}
}
}
if (count1 >= count2)
System.out.println(true);
else
System.out.println(false);
}

Java compressing Strings

I need to create a method that receives a String and also returns a String.
Ex input: AAABBBBCC
Ex output: 3A4B2C
Well, this is quite embarrassing and I couldn't manage to do it on the interview that I had today ( I was applying for a Junior position ), now, trying at home I made something that works statically, I mean, not using a loop which is kind of useless but I don't know if I'm not getting enough hours of sleep or something but I can't figure it out how my for loop should look like. This is the code:
public static String Comprimir(String texto){
StringBuilder objString = new StringBuilder();
int count;
char match;
count = texto.substring(texto.indexOf(texto.charAt(1)), texto.lastIndexOf(texto.charAt(1))).length()+1;
match = texto.charAt(1);
objString.append(count);
objString.append(match);
return objString.toString();
}
Thanks for your help, I'm trying to improve my logic skills.

Loop through the string remembering what you last saw. Every time you see the same letter count. When you see a new letter put what you have counted onto the output and set the new letter as what you have last seen.
String input = "AAABBBBCC";
int count = 1;
char last = input.charAt(0);
StringBuilder output = new StringBuilder();
for(int i = 1; i < input.length(); i++){
if(input.charAt(i) == last){
count++;
}else{
if(count > 1){
output.append(""+count+last);
}else{
output.append(last);
}
count = 1;
last = input.charAt(i);
}
}
if(count > 1){
output.append(""+count+last);
}else{
output.append(last);
}
System.out.println(output.toString());

You can do that using the following steps:
Create a HashMap
For every character, Get the value from the hashmap
-If the value is null, enter 1
-else, replace the value with (value+1)
Iterate over the HashMap and keep concatenating (Value+Key)

use StringBuilder (you did that)
define two variables - previousChar and counter
loop from 0 to str.length() - 1
each time get str.charat(i) and compare it to what's stored in the previousChar variable
if the previous char is the same, increment a counter
if the previous char is not the same, and counter is 1, increment counter
if the previous char is not the same, and counter is >1, append counter + currentChar, reset counter
after the comparison, assign the current char previousChar
cover corner cases like "first char"
Something like that.

The easiest approach:- Time Complexity - O(n)
public static void main(String[] args) {
String str = "AAABBBBCC"; //input String
int length = str.length(); //length of a String
//Created an object of a StringBuilder class
StringBuilder sb = new StringBuilder();
int count=1; //counter for counting number of occurances
for(int i=0; i<length; i++){
//if i reaches at the end then append all and break the loop
if(i==length-1){
sb.append(str.charAt(i)+""+count);
break;
}
//if two successive chars are equal then increase the counter
if(str.charAt(i)==str.charAt(i+1)){
count++;
}
else{
//else append character with its count
sb.append(str.charAt(i)+""+count);
count=1; //reseting the counter to 1
}
}
//String representation of a StringBuilder object
System.out.println(sb.toString());
}

In the count=... line, lastIndexOf will not care about consecutive values, and will just give the last occurence.
For instance, in the string "ABBA", the substring would be the whole string.
Also, taking the length of the substring is equivalent to subtracting the two indexes.
I really think that you need a loop.
Here is an example :
public static String compress(String text) {
String result = "";
int index = 0;
while (index < text.length()) {
char c = text.charAt(index);
int count = count(text, index);
if (count == 1)
result += "" + c;
else
result += "" + count + c;
index += count;
}
return result;
}
public static int count(String text, int index) {
char c = text.charAt(index);
int i = 1;
while (index + i < text.length() && text.charAt(index + i) == c)
i++;
return i;
}
public static void main(String[] args) {
String test = "AAABBCCC";
System.out.println(compress(test));
}

Please try this one. This may help to print the count of characters which we pass on string format through console.
import java.util.*;
public class CountCharacterArray {
private static Scanner inp;
public static void main(String args[]) {
inp = new Scanner(System.in);
String str=inp.nextLine();
List<Character> arrlist = new ArrayList<Character>();
for(int i=0; i<str.length();i++){
arrlist.add(str.charAt(i));
}
for(int i=0; i<str.length();i++){
int freq = Collections.frequency(arrlist, str.charAt(i));
System.out.println("Frequency of "+ str.charAt(i)+ " is: "+freq);
}
}
}

Java's not my main language, hardly ever use it, but I wanted to give it a shot :]
Not even sure if your assignment requires a loop, but here's a regexp approach:
public static String compress_string(String inp) {
String compressed = "";
Pattern pattern = Pattern.compile("([\\w])\\1*");
Matcher matcher = pattern.matcher(inp);
while(matcher.find()) {
String group = matcher.group();
if (group.length() > 1) compressed += group.length() + "";
compressed += group.charAt(0);
}
return compressed;
}

This is just one more way of doing it.
public static String compressor(String raw) {
StringBuilder builder = new StringBuilder();
int counter = 0;
int length = raw.length();
int j = 0;
while (counter < length) {
j = 0;
while (counter + j < length && raw.charAt(counter + j) == raw.charAt(counter)) {
j++;
}
if (j > 1) {
builder.append(j);
}
builder.append(raw.charAt(counter));
counter += j;
}
return builder.toString();
}

The following can be used if you are looking for a basic solution. Iterate through the string with one element and after finding all the element occurrences, remove that character. So that it will not interfere in the next search.
public static void main(String[] args) {
String string = "aaabbbbbaccc";
int counter;
String result="";
int i=0;
while (i<string.length()){
counter=1;
for (int j=i+1;j<string.length();j++){
System.out.println("string length ="+string.length());
if (string.charAt(i) == string.charAt(j)){
counter++;
}
}
result = result+string.charAt(i)+counter;
string = string.replaceAll(String.valueOf(string.charAt(i)), "");
}
System.out.println("result is = "+result);
}
And the output will be :=
result is = a4b5c3

private String Comprimir(String input){
String output="";
Map<Character,Integer> map=new HashMap<Character,Integer>();
for(int i=0;i<input.length();i++){
Character character=input.charAt(i);
if(map.containsKey(character)){
map.put(character, map.get(character)+1);
}else
map.put(character, 1);
}
for (Entry<Character, Integer> entry : map.entrySet()) {
output+=entry.getValue()+""+entry.getKey().charValue();
}
return output;
}
One other simple way using Multiset of guava-
import java.util.Arrays;
import com.google.common.collect.HashMultiset;
import com.google.common.collect.Multiset;
import com.google.common.collect.Multiset.Entry;
public class WordSpit {
public static void main(String[] args) {
String output="";
Multiset<String> wordsMultiset = HashMultiset.create();
String[] words="AAABBBBCC".split("");
wordsMultiset.addAll(Arrays.asList(words));
for (Entry<String> string : wordsMultiset.entrySet()) {
if(!string.getElement().isEmpty())
output+=string.getCount()+""+string.getElement();
}
System.out.println(output);
}
}

consider the below Solution in which the String s1 identifies the unique characters that are available in a given String s (for loop 1), in the second for loop build a string s2 that contains unique character and no of times it is repeated by comparing string s1 with s.
public static void main(String[] args)
{
// TODO Auto-generated method stub
String s = "aaaabbccccdddeee";//given string
String s1 = ""; // string to identify how many unique letters are available in a string
String s2=""; //decompressed string will be appended to this string
int count=0;
for(int i=0;i<s.length();i++) {
if(s1.indexOf(s.charAt(i))<0) {
s1 = s1+s.charAt(i);
}
}
for(int i=0;i<s1.length();i++) {
for(int j=0;j<s.length();j++) {
if(s1.charAt(i)==s.charAt(j)) {
count++;
}
}
s2=s2+s1.charAt(i)+count;
count=0;
}
System.out.println(s2);
}

It may help you.
public class StringCompresser
{
public static void main(String[] args)
{
System.out.println(compress("AAABBBBCC"));
System.out.println(compress("AAABC"));
System.out.println(compress("A"));
System.out.println(compress("ABBDCC"));
System.out.println(compress("AZXYC"));
}
static String compress(String str)
{
StringBuilder stringBuilder = new StringBuilder();
char[] charArray = str.toCharArray();
int count = 1;
char lastChar = 0;
char nextChar = 0;
lastChar = charArray[0];
for (int i = 1; i < charArray.length; i++)
{
nextChar = charArray[i];
if (lastChar == nextChar)
{
count++;
}
else
{
stringBuilder.append(count).append(lastChar);
count = 1;
lastChar = nextChar;
}
}
stringBuilder.append(count).append(lastChar);
String compressed = stringBuilder.toString();
return compressed;
}
}
Output:
3A4B2C
3A1B1C
1A
1A2B1D2C
1A1Z1X1Y1C

The answers which used Map will not work for cases like aabbbccddabc as in that case the output should be a2b3c2d2a1b1c1.
In that case this implementation can be used :
private String compressString(String input) {
String output = "";
char[] arr = input.toCharArray();
Map<Character, Integer> myMap = new LinkedHashMap<>();
for (int i = 0; i < arr.length; i++) {
if (i > 0 && arr[i] != arr[i - 1]) {
output = output + arr[i - 1] + myMap.get(arr[i - 1]);
myMap.put(arr[i - 1], 0);
}
if (myMap.containsKey(arr[i])) {
myMap.put(arr[i], myMap.get(arr[i]) + 1);
} else {
myMap.put(arr[i], 1);
}
}
for (Character c : myMap.keySet()) {
if (myMap.get(c) != 0) {
output = output + c + myMap.get(c);
}
}
return output;
}

O(n) approach
No need for hashing. The idea is to find the first Non-matching character.
The count of each character would be the difference in the indices of both characters.
for a detailed answer: https://stackoverflow.com/a/55898810/7972621
The only catch is that we need to add a dummy letter so that the comparison for
the last character is possible.
private static String compress(String s){
StringBuilder result = new StringBuilder();
int j = 0;
s = s + '#';
for(int i=1; i < s.length(); i++){
if(s.charAt(i) != s.charAt(j)){
result.append(i-j);
result.append(s.charAt(j));
j = i;
}
}
return result.toString();
}

The code below will ask the user for user to input a specific character to count the occurrence .
import java.util.Scanner;
class CountingOccurences {
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();
System.out.println("Enter th Char to see the occurence\n");
ch=inp.next().charAt(0);
for(int i=0;i<str.length();i++)
{
if(str.charAt(i)==ch)
{
count++;
}
}
System.out.println("The Character is Occuring");
System.out.println(count+"Times");
}
}

public static char[] compressionTester( char[] s){
if(s == null){
throw new IllegalArgumentException();
}
HashMap<Character, Integer> map = new HashMap<>();
for (int i = 0 ; i < s.length ; i++) {
if(!map.containsKey(s[i])){
map.put(s[i], 1);
}
else{
int value = map.get(s[i]);
value++;
map.put(s[i],value);
}
}
String newer="";
for( Character n : map.keySet()){
newer = newer + n + map.get(n);
}
char[] n = newer.toCharArray();
if(s.length > n.length){
return n;
}
else{
return s;
}
}

package com.tell.datetime;
import java.util.Stack;
public class StringCompression {
public static void main(String[] args) {
String input = "abbcccdddd";
System.out.println(compressString(input));
}
public static String compressString(String input) {
if (input == null || input.length() == 0)
return input;
String finalCompressedString = "";
String lastElement="";
char[] charArray = input.toCharArray();
Stack stack = new Stack();
int elementCount = 0;
for (int i = 0; i < charArray.length; i++) {
char currentElement = charArray[i];
if (i == 0) {
stack.push((currentElement+""));
continue;
} else {
if ((currentElement+"").equalsIgnoreCase((String)stack.peek())) {
stack.push(currentElement + "");
if(i==charArray.length-1)
{
while (!stack.isEmpty()) {
lastElement = (String)stack.pop();
elementCount++;
}
finalCompressedString += lastElement + "" + elementCount;
}else
continue;
}
else {
while (!stack.isEmpty()) {
lastElement = (String)stack.pop();
elementCount++;
}
finalCompressedString += lastElement + "" + elementCount;
elementCount=0;
stack.push(currentElement+"");
}
}
}
if (finalCompressedString.length() >= input.length())
return input;
else
return finalCompressedString;
}
}

public class StringCompression {
public static void main(String[] args){
String s = "aabcccccaaazdaaa";
char check = s.charAt(0);
int count = 0;
for(int i=0; i<s.length(); i++){
if(s.charAt(i) == check) {
count++;
if(i==s.length()-1){
System.out.print(s.charAt(i));
System.out.print(count);
}
} else {
System.out.print(s.charAt(i-1));
System.out.print(count);
check = s.charAt(i);
count = 1;
if(i==s.length()-1){
System.out.print(s.charAt(i));
System.out.print(count);
}
}
}
}

// O(N) loop through entire character array
// match current char with next one, if they matches count++
// if don't then just append current char and counter value and then reset counter.
// special case is the last characters, for that just check if count value is > 0, if it's then append the counter value and the last char
private String compress(String str) {
char[] c = str.toCharArray();
String newStr = "";
int count = 1;
for (int i = 0; i < c.length - 1; i++) {
int j = i + 1;
if (c[i] == c[j]) {
count++;
} else {
newStr = newStr + c[i] + count;
count = 1;
}
}
// this is for the last strings...
if (count > 0) {
newStr = newStr + c[c.length - 1] + count;
}
return newStr;
}

public class StringCompression {
public static void main(String... args){
String s="aabbcccaa";
//a2b2c3a2
for(int i=0;i<s.length()-1;i++){
int count=1;
while(i<s.length()-1 && s.charAt(i)==s.charAt(i+1)){
count++;
i++;
}
System.out.print(s.charAt(i));
System.out.print(count);
}
System.out.println(" ");
}
}

This is a leet code problem 443. Most of the answers here uses StringBuilder or a HashMap, the actual problem statement is to solve using the input char array and in place array modification.
public int compress(char[] chars) {
int startIndex = 0;
int lastArrayIndex = 0;
if (chars.length == 1) {
return 1;
}
if (chars.length == 0) {
return 0;
}
for (int j = startIndex + 1; j < chars.length; j++) {
if (chars[startIndex] != chars[j]) {
chars[lastArrayIndex] = chars[startIndex];
lastArrayIndex++;
if ((j - startIndex) > 1) {
for (char c : String.valueOf(j - startIndex).toCharArray()) {
chars[lastArrayIndex] = c;
lastArrayIndex++;
}
}
startIndex = j;
}
if (j == chars.length - 1) {
if (j - startIndex >= 1) {
j = chars.length;
chars[lastArrayIndex] = chars[startIndex];
lastArrayIndex++;
for (char c : String.valueOf(j - startIndex).toCharArray()) {
chars[lastArrayIndex] = c;
lastArrayIndex++;
}
} else {
chars[lastArrayIndex] = chars[startIndex];
lastArrayIndex++;
}
}
}
return lastArrayIndex;
}
}

Get the longest substring from a string which is passed in Java [closed]

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I want to pass in a string into a method e.g. "abbcccdef" and want it to return the longest substring. in this case it would be "ccc". Someone please help me with code that would help me solve this issue. I would like something basic that would allow a beginner to understand easily. This is what i have so far but it doesnt seem to work:
Many Thanks
public String getLongestSubstring(String s) {
int [] length = new int [s.length()];
String longestString = "";
if (s.length() > 0) {
char c = s.charAt(0);
for (int i=0;i<s.length();i++) {
for (int j=0;j<s.length();j++) {
if (c==s.charAt(j)) {
length [i]++;
} else {
c = s.charAt(j);
i++;
}
}
}
return longestString;
}
else
return null;
}

Follow the bellow mentioned algorithm sub-string sequence

I haven't tested this and I found and fixed one error already, but I think I've covered all the bases now. One limitation you didn't elaborate on in your question was the case in which there are two substrings of equal, longest length. i.e. abbbcccdef, I just return the first.
public String getLongestSubstring(String s)
{
if (s.length() == 0) return null;
//We need some data to start with.
char currentChar = s.charAt(0);
String longestString = "" + currentChar;
String runningString = "" + currentChar;
int currentLongestLength = 1;
for (int i = 1; i < s.length(); i++)
{
//Check the current char, is it the same?
if (s.charAt(i) == currentChar)
{
runningString = runningString + currentChar;
//Have we beaten our previous record.
if (runningString.length() > longestString.length())
{
//Increase the record.
currentLongestLength++;
longestString = runningString;
}
}
else
{
runningString = "" + s.charAt(i);
}
}
return longestString;
}

Here is how I implemented to find a longest SubString from a String.
public class LongestString {
public static void main (String[] args) throws java.lang.Exception {
System.out.println("Longest Substring is " + getLongestSubstring("abbcccdf"));
}
public static String getLongestSubstring(String s) {
int length = 1;
String longestString = "";
for (int i = 0; i < s.length(); i++) {
StringBuilder str = new StringBuilder();
str.append(s.charAt(i));
for (int j = i + 1; j < s.length(); j++) {
if (s.charAt(i) == s.charAt(j)) {
str.append(s.charAt(j));
} else {
break;
}
}
if (length < str.length()) {
length = str.length();
longestString = str.toString();
}
}
return longestString;
}
}
Test it: http://ideone.com/JM904Y

public String getLongestSubString(String source) {
StringBuilder longestStringBuilder = new StringBuilder(source.length());
String longestString = "";
for(int i=0; i<source.length(); i++) {
if(longestStringBuilder.toString().contains(String.valueOf(source.charAt(i)))) {
if(longestString.length() < longestStringBuilder.toString().length()) {
longestString = longestStringBuilder.toString();
}
longestStringBuilder.setLength(0);
}
longestStringBuilder.append(source.charAt(i));
}
if(longestString.length() < longestStringBuilder.toString().length()) {
longestString = longestStringBuilder.toString();
}
return longestString;
}