Develop Reference

Problem with Java code that will print TRUE when a word has no vowels in an IF statement

I am trying to write a code the will print TRUE when i enter a word with no vowels and will print FALSE if I enter a word with vowels. I am having trouble when it comes to writing the string. I give one of the strings the value that holds all the values.
String A = "aeiou";
The other 2 strings hold the rest of the alphabet.
When I type in a word that has vocals it prints FALSE however when I type in a word with no vowels it prints FALSE.
I wanna know if there is a way to make the program read the values of the string as separate entities and not as a whole line.
package lab3;
import java.util.Scanner;
public class Lab3 {
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
String input = reader.next();
/* A and B, but not C */
String A = "bcdfghjklm";
String B = "npqrstvwxyz";
String C = "aeiou";
if (input.contains(A) && input.contains(B) || input.contains(C)) {
System.out.println("TRUE");
} else {
System.out.println("FALSE");
}
}
}

You can do it with a short function
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
String input = reader.next();
System.out.println(chk(input));
}
static public boolean chk(String s) {
for (int i = 0; i < s.length(); i++) {
if ("aeiou".indexOf(s.charAt(i)) != -1) {
return false;
}
}
return true;
}
}

This is a working correct for your purpose:
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
String input = reader.next();
String C = "aeiou";
for (int i = 0; i < input.length(); i++) {
char inputA = input.charAt(i);
for (int j = 0; j < C.length(); j++) {
char c = C.charAt(j);
if (inputA == c) {
System.out.println("TRUE");
return;
}
}
}
System.out.println("FALSE");
}
The key point is to use the method charAt(int index); so you can check each char in the string.

You can choose to use any of the below two methods, containsVowel and containsVowel1 as both of them do the same thing:
import java.util.Arrays;
import java.util.Scanner;
public class Test {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
System.out.println("Enter a text: ");
String input = reader.nextLine();
String[] vowels = { "a", "e", "i", "o", "u" };
System.out.println(!containsVowel(input, vowels));
System.out.println(!containsVowel1(input, vowels));
}
public static boolean containsVowel(String input, String[] vowels) {
return Arrays.stream(vowels).parallel().anyMatch(input::contains);
}
public static boolean containsVowel1(String input, String[] vowels) {
for (int i = 0; i < vowels.length; i++) {
if (input.contains(vowels[i])) {
return true;
}
}
return false;
}
}

Java: Print a unique character in a string

I'm writing a program that will print the unique character in a string (entered through a scanner). I've created a method that tries to accomplish this but I keep getting characters that are not repeats, instead of a character (or characters) that is unique to the string. I want the unique letters only.
Here's my code:
import java.util.Scanner;
public class Sameness{
public static void main (String[]args){
Scanner kb = new Scanner (System.in);
String word = "";
System.out.println("Enter a word: ");
word = kb.nextLine();
uniqueCharacters(word);
}
public static void uniqueCharacters(String test){
String temp = "";
for (int i = 0; i < test.length(); i++){
if (temp.indexOf(test.charAt(i)) == - 1){
temp = temp + test.charAt(i);
}
}
System.out.println(temp + " ");
}
}
And here's sample output with the above code:
Enter a word:
nreena
nrea
The expected output would be: ra

Based on your desired output, you have to replace a character that initially has been already added when it has a duplicated later, so:
public static void uniqueCharacters(String test){
String temp = "";
for (int i = 0; i < test.length(); i++){
char current = test.charAt(i);
if (temp.indexOf(current) < 0){
temp = temp + current;
} else {
temp = temp.replace(String.valueOf(current), "");
}
}
System.out.println(temp + " ");
}

How about applying the KISS principle:
public static void uniqueCharacters(String test) {
System.out.println(test.chars().distinct().mapToObj(c -> String.valueOf((char)c)).collect(Collectors.joining()));
}

The accepted answer will not pass all the test case for example
input -"aaabcdd"
desired output-"bc"
but the accepted answer will give -abc
because the character a present odd number of times.
Here I have used ConcurrentHasMap to store character and the number of occurrences of character then removed the character if the occurrences is more than one time.
import java.util.concurrent.ConcurrentHashMap;
public class RemoveConductive {
public static void main(String[] args) {
String s="aabcddkkbghff";
String[] cvrtar=s.trim().split("");
ConcurrentHashMap<String,Integer> hm=new ConcurrentHashMap<>();
for(int i=0;i<cvrtar.length;i++){
if(!hm.containsKey(cvrtar[i])){
hm.put(cvrtar[i],1);
}
else{
hm.put(cvrtar[i],hm.get(cvrtar[i])+1);
}
}
for(String ele:hm.keySet()){
if(hm.get(ele)>1){
hm.remove(ele);
}
}
for(String key:hm.keySet()){
System.out.print(key);
}
}
}

Though to approach a solution I would suggest you to try and use a better data structure and not just string. Yet, you can simply modify your logic to delete already existing duplicates using an else as follows :
public static void uniqueCharacters(String test) {
String temp = "";
for (int i = 0; i < test.length(); i++) {
char ch = test.charAt(i);
if (temp.indexOf(ch) == -1) {
temp = temp + ch;
} else {
temp.replace(String.valueOf(ch),""); // added this to your existing code
}
}
System.out.println(temp + " ");
}

This is an interview question. Find Out all the unique characters of a string.
Here is the complete solution. The code itself is self explanatory.
public class Test12 {
public static void main(String[] args) {
String a = "ProtijayiGiniGina";
allunique(a);
}
private static void allunique(String a) {
int[] count = new int[256];// taking count of characters
for (int i = 0; i < a.length(); i++) {
char ch = a.charAt(i);
count[ch]++;
}
for (int i = 0; i < a.length(); i++) {
char chh = a.charAt(i);
// character which has arrived only one time in the string will be printed out
if (count[chh] == 1) {
System.out.println("index => " + i + " and unique character => " + a.charAt(i));
}
}
}// unique
}
In Python :
def firstUniqChar(a):
count = [0] *256
for i in a: count[ord(i)] += 1
element = ""
for item in a:
if (count[ord(item)] == 1):
element = item;
break;
return element
a = "GiniGinaProtijayi";
print(firstUniqChar(a)) # output is P

public static String input = "10 5 5 10 6 6 2 3 1 3 4 5 3";
public static void uniqueValue (String numbers) {
String [] str = input.split(" ");
Set <String> unique = new HashSet <String> (Arrays.asList(str));
System.out.println(unique);
for (String value:unique) {
int count = 0;
for ( int i= 0; i<str.length; i++) {
if (value.equals(str[i])) {
count++;
}
}
System.out.println(value+"\t"+count);
}
}
public static void main(String [] args) {
uniqueValue(input);
}

Step1: To find the unique characters in a string, I have first taken the string from user.
Step2: Converted the input string to charArray using built in function in java.
Step3: Considered two HashSet (set1 for storing all characters even if it is getting repeated, set2 for storing only unique characters.
Step4 : Run for loop over the array and check that if particular character is not there in set1 then add it to both set1 and set2. if that particular character is already there in set1 then add it to set1 again but remove it from set2.( This else part is useful when particular character is getting repeated odd number of times).
Step5 : Now set2 will have only unique characters. Hence, just print that set2.
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
String str = input.next();
char arr[] = str.toCharArray();
HashSet<Character> set1=new HashSet<Character>();
HashSet<Character> set2=new HashSet<Character>();
for(char i:arr)
{
if(set1.contains(i))
{
set1.add(i);
set2.remove(i);
}
else
{
set1.add(i);
set2.add(i);
}
}
System.out.println(set2);
}

I would store all the characters of the string in an array that you will loop through to check if the current characters appears there more than once. If it doesn't, then add it to temp.
public static void uniqueCharacters(String test) {
String temp = "";
char[] array = test.toCharArray();
int count; //keep track of how many times the character exists in the string
outerloop: for (int i = 0; i < test.length(); i++) {
count = 0; //reset the count for every new letter
for(int j = 0; j < array.length; j++) {
if(test.charAt(i) == array[j])
count++;
if(count == 2){
count = 0;
continue outerloop; //move on to the next letter in the string; this will skip the next two lines below
}
}
temp += test.charAt(i);
System.out.println("Adding.");
}
System.out.println(temp);
}
I have added comments for some more detail.

import java.util.*;
import java.lang.*;
class Demo
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter String");
String s1=sc.nextLine();
try{
HashSet<Object> h=new HashSet<Object>();
for(int i=0;i<s1.length();i++)
{
h.add(s1.charAt(i));
}
Iterator<Object> itr=h.iterator();
while(itr.hasNext()){
System.out.println(itr.next());
}
}
catch(Exception e)
{
System.out.println("error");
}
}
}

If you don't want to use additional space:
String abc="developer";
System.out.println("The unique characters are-");
for(int i=0;i<abc.length();i++)
{
for(int j=i+1;j<abc.length();j++)
{
if(abc.charAt(i)==abc.charAt(j))
abc=abc.replace(String.valueOf(abc.charAt(j))," ");
}
}
System.out.println(abc);
Time complexity O(n^2) and no space.

This String algorithm is used to print unique characters in a string.It runs in O(n) runtime where n is the length of the string.It supports ASCII characters only.
static String printUniqChar(String s) {
StringBuilder buildUniq = new StringBuilder();
boolean[] uniqCheck = new boolean[128];
for (int i = 0; i < s.length(); i++) {
if (!uniqCheck[s.charAt(i)]) {
uniqCheck[s.charAt(i)] = true;
if (uniqCheck[s.charAt(i)])
buildUniq.append(s.charAt(i));
}
}

public class UniqueCharactersInString {
public static void main(String []args){
String input = "aabbcc";
String output = uniqueString(input);
System.out.println(output);
}
public static String uniqueString(String s){
HashSet<Character> uniques = new HashSet<>();
uniques.add(s.charAt(0));
String out = "";
out += s.charAt(0);
for(int i =1; i < s.length(); i++){
if(!uniques.contains(s.charAt(i))){
uniques.add(s.charAt(i));
out += s.charAt(i);
}
}
return out;
}
}
What would be the inneficiencies of this answer? How does it compare to other answers?

Based on your desired output you can replace each character already present with a blank character.
public static void uniqueCharacters(String test){
String temp = "";
for(int i = 0; i < test.length(); i++){
if (temp.indexOf(test.charAt(i)) == - 1){
temp = temp + test.charAt(i);
} else {
temp.replace(String.valueOf(temp.charAt(i)), "");
}
}
System.out.println(temp + " ");
}

public void uniq(String inputString) {
String result = "";
int inputStringLen = inputStr.length();
int[] repeatedCharacters = new int[inputStringLen];
char inputTmpChar;
char tmpChar;
for (int i = 0; i < inputStringLen; i++) {
inputTmpChar = inputStr.charAt(i);
for (int j = 0; j < inputStringLen; j++) {
tmpChar = inputStr.charAt(j);
if (inputTmpChar == tmpChar)
repeatedCharacters[i]++;
}
}
for (int k = 0; k < inputStringLen; k++) {
inputTmpChar = inputStr.charAt(k);
if (repeatedCharacters[k] == 1)
result = result + inputTmpChar + " ";
}
System.out.println ("Unique characters: " + result);
}
In first for loop I count the number of times the character repeats in the string. In the second line I am looking for characters repetitive once.

how about this :)
for (int i=0; i< input.length();i++)
if(input.indexOf(input.charAt(i)) == input.lastIndexOf(input.charAt(i)))
System.out.println(input.charAt(i) + " is unique");

package extra;
public class TempClass {
public static void main(String[] args) {
// TODO Auto-generated method stub
String abcString="hsfj'pwue2hsu38bf74sa';fwe'rwe34hrfafnosdfoasq7433qweid";
char[] myCharArray=abcString.toCharArray();
TempClass mClass=new TempClass();
mClass.countUnique(myCharArray);
mClass.countEach(myCharArray);
}
/**
* This is the program to find unique characters in array.
* #add This is nice.
* */
public void countUnique(char[] myCharArray) {
int arrayLength=myCharArray.length;
System.out.println("Array Length is: "+arrayLength);
char[] uniqueValues=new char[myCharArray.length];
int uniqueValueIndex=0;
int count=0;
for(int i=0;i<arrayLength;i++) {
for(int j=0;j<arrayLength;j++) {
if (myCharArray[i]==myCharArray[j] && i!=j) {
count=count+1;
}
}
if (count==0) {
uniqueValues[uniqueValueIndex]=myCharArray[i];
uniqueValueIndex=uniqueValueIndex+1;
count=0;
}
count=0;
}
for(char a:uniqueValues) {
System.out.println(a);
}
}
/**
* This is the program to find count each characters in array.
* #add This is nice.
* */
public void countEach(char[] myCharArray) {
}
}

Here str will be your string to find the unique characters.
function getUniqueChars(str){
let uniqueChars = '';
for(let i = 0; i< str.length; i++){
for(let j= 0; j< str.length; j++) {
if(str.indexOf(str[i]) === str.lastIndexOf(str[j])) {
uniqueChars += str[i];
}
}
}
return uniqueChars;
}

public static void main(String[] args) {
String s = "aaabcdd";
char a[] = s.toCharArray();
List duplicates = new ArrayList();
List uniqueElements = new ArrayList();
for (int i = 0; i < a.length; i++) {
uniqueElements.add(a[i]);
for (int j = i + 1; j < a.length; j++) {
if (a[i] == a[j]) {
duplicates.add(a[i]);
break;
}
}
}
uniqueElements.removeAll(duplicates);
System.out.println(uniqueElements);
System.out.println("First Unique : "+uniqueElements.get(0));
}
Output :
[b, c]
First Unique : b

import java.util.*;
public class Sameness{
public static void main (String[]args){
Scanner kb = new Scanner (System.in);
String word = "";
System.out.println("Enter a word: ");
word = kb.nextLine();
uniqueCharacters(word);
}
public static void uniqueCharacters(String test){
for(int i=0;i<test.length();i++){
if(test.lastIndexOf(test.charAt(i))!=i)
test=test.replaceAll(String.valueOf(test.charAt(i)),"");
}
System.out.println(test);
}
}

public class Program02
{
public static void main(String[] args)
{
String inputString = "abhilasha";
for (int i = 0; i < inputString.length(); i++)
{
for (int j = i + 1; j < inputString.length(); j++)
{
if(inputString.toCharArray()[i] == inputString.toCharArray()[j])
{
inputString = inputString.replace(String.valueOf(inputString.charAt(j)), "");
}
}
}
System.out.println(inputString);
}
}

The requirement is, the output will print one number followed by a character then a number and so on

I have am assignment suppose there is a string 11234aBcD the out put should be 1a1B2c3D4 and I am unable to do it, First and the second output I have done. And I have to do using java.
I am adding my code:
package Testx;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
public class Test3optmz {
public static void main(String[] args)
{
// TODO Auto-generated method stub
Scanner gaba = new Scanner( System.in );
String variable;
System.out.print("Enter String:");
variable = gaba.nextLine();
SeparateGaba(variable);
}
public static void SeparateGaba(String str)
{
String Catch_num = "";
String Catch_let = "";
String upper="";
String lower="";
String holdl="";
String holdn="";
for (int i = 0; i < str.length(); i++)
{
char a = str.charAt(i);
if (Character.isDigit(a))
{
Catch_num = Catch_num + a;
} else
{
Catch_let = Catch_let + a;
}
}
System.out.println("FIRST OUTPUT:"+Catch_num+Catch_let);// i am separating the numbers and alphabets
for(int j=0;j <Catch_let.length();j++)
{
char x = Catch_let.charAt(j);
if (Character.isUpperCase(x))
{
upper += x;
}
else
{
lower += x;
}
}
char[] num = Catch_num.toCharArray();
Arrays.sort(num);
String n =new String(num);
char[] ordr = lower.toCharArray();
Arrays.sort(ordr);
String alfa1 =new String(ordr);
char[] ord = upper.toCharArray();
Arrays.sort(ord);
String alfa =new String(ord);
String t= alfa1 + alfa;
char [] fin = t.toCharArray();
Arrays.sort(fin);
String fal = new String(fin);
Character[] chars = new Character[fal.length()];
for (int f=0; f < fal.length();f++)
chars[f] = fal.charAt(f);
Arrays.sort(chars, new Comparator<Character>()
{
public int compare(Character c1, Character c2)
{
int cmp = Character.compare(
Character.toLowerCase(c1.charValue()),
Character.toLowerCase(c2.charValue())
);
if (cmp != 0) return cmp;
return Character.compare(c1.charValue(), c2.charValue());
}
});
StringBuilder sb = new StringBuilder(chars.length);
for (char c : chars) sb.append(c);
fal = sb.toString();
System.out.println("SECOND OUTPUT= "+n+fal);// now i am arranging the alphabet in ascending order avoiding capital and small ex 1234eBaC it will come as 1234aBce
}
}

well thanks for not helping me out i helped my self
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;
public class Test4 {
public static void main(String[] args)
{
// TODO Auto-generated method stub
Scanner gaba = new Scanner( System.in );
String variable;
System.out.print("Enter String:");
variable = gaba.nextLine();
SeparateGaba(variable);
}
public static void SeparateGaba(String str)
{
String Catch_num = "";
String Catch_let = "";
String upper="";
String lower="";
for (int i = 0; i < str.length(); i++)
{
char a = str.charAt(i);
if (Character.isDigit(a))
{
Catch_num = Catch_num + a;
} else
{
Catch_let = Catch_let + a;
}
}
System.out.println("FIRST OUTPUT:"+Catch_num+Catch_let);
for(int j=0;j <Catch_let.length();j++)
{
char x = Catch_let.charAt(j);
if (Character.isUpperCase(x))
{
upper += x;
}
else
{
lower += x;
}
}
char[] num = Catch_num.toCharArray();
Arrays.sort(num);
String n =new String(num);
char[] ordr = lower.toCharArray();
Arrays.sort(ordr);
String alfa1 =new String(ordr);
char[] ord = upper.toCharArray();
Arrays.sort(ord);
String alfa =new String(ord);
String t= alfa1 + alfa;
char [] fin = t.toCharArray();
Arrays.sort(fin);
String fal = new String(fin);
Character[] chars = new Character[fal.length()];
for (int f=0; f < fal.length();f++)
chars[f] = fal.charAt(f);
Arrays.sort(chars, new Comparator<Character>()
{
public int compare(Character c1, Character c2)
{
int cmp = Character.compare(
Character.toLowerCase(c1.charValue()),
Character.toLowerCase(c2.charValue())
);
if (cmp != 0) return cmp;
return Character.compare(c1.charValue(), c2.charValue());
}
});
StringBuilder sb = new StringBuilder(chars.length);
for (char c : chars) sb.append(c);
fal = sb.toString();
System.out.println("SECOND OUTPUT= "+n+fal);
/*GABA come on let seeeeeeeee*/
try
{
for(int i=0;i<n.length();i++)
{
char a= n.charAt(i);
System.out.print(a);
for(int j=0; j<fal.length();j++)//1
{
char b = fal.charAt(j);
if(j>=i)
{
System.out.print(b);
break;
}
}
}
}
catch(Exception e)
{
System.out.println("Your letter length should match with char len");
}
}
}
Enter String:14a5Bc
FIRST OUTPUT:145aBc
SECOND OUTPUT= 145aBc
THE out put i was looking for --->1a4B5c

Reverse word letters in a sentence java

For my program I want to reverse letters in a line of text. However I don't want to reverse the order of the words in the sentence.
For example when I input: "This is a string"
I get: gnirts a si siht
But I want: siht si a gnirts
public static String reverseWordCharacters(String text1) {
String reverse = "";
int length = text1.length();
for (int i = length - 1; i >= 0; i--) {
reverse = reverse + text1.charAt(i);
System.out.println();
}
return reverse;
}
}

String sentence = "This is a string";
String[] words = sentence.split(" ");
String invertedSentece = "";
for (String word : words){
String invertedWord = "";
for (int i = word.length() - 1; i >= 0; i--)
invertedWord += word.charAt(i);
invertedSentece += invertedWord;
invertedSentece += " ";
}
invertedSentece.trim();

Commons Lang's StringUtils class has method to do this:
http://commons.apache.org/proper/commons-lang/javadocs/api-release/org/apache/commons/lang3/StringUtils.html#reverse(java.lang.String)

This may be the easiest way
import java.util.*;
public class ReverseString{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
System.out.print("Enter any Sentence: ");
String sentence = sc.nextLine();
String[] stringArray = sentence.split(" ");
for(int i = 0; i<stringArray.length; i++){
String temp = new StringBuilder(stringArray[i]).reverse().toString();
System.out.print(temp+" ");
}
}
}

private static String revWrord(String s) {
LinkedList word = new LinkedList();
Scanner scanfs = new Scanner(s);
while(scanfs.hasNext()) {
word.add(scanfs.next());
}
StringBuilder output = new StringBuilder();
StringBuilder temp = new StringBuilder();
while(true) {
temp.append(word.poll()).reverse();
output.append(temp);
temp.delete(0,temp.length() );
if(word.isEmpty()) {
break;
} else {
output.append(" ");
}
}
return output.toString();
}

public static void main(String[] args) {
String s= "This is a string";
System.out.println(s);
String finalrevresestring = "";
String wordreverse = "";
String[] a = s.split(" ");
for (int i = 0; i<=a.length-1; i++) {
for (int x=a[i].length()-1; x>=0; x--) {
wordreverse += a[i].charAt(x);
}
finalrevresestring += wordreverse;
finalrevresestring += " ";
wordreverse = "";
}
System.out.println("This is reverse: "+finalrevresestring);
}

{
String sentence="This is String";
String[] words=sentence.split(" ");
String inverted=" ";
for(String eachWord : words){
for(int i=0;i<eachWord.length();i++){
inverted+=eachWord.charAt(eachWord.length()-1-i);
}
inverted+=" ";
}
System.out.println("Inverted words in the given sentence is :"+inverted);
}

StringBuffer s=new StringBuffer("I live in India");
int i=0,j=0,w=0;
//w is first position of word
//j is last position of word
while(i<s.lenght){
if(s.charAt[i]==" "||i==s.lenght()-1)
{
j=i;
if(i==s.length-1)
j=i;
else
j--;
swapMethod(w,j);
w=i+1;
}
i++;
}

I want to split string without using split function?

I want to split string without using split . can anybody solve my problem I am tried but
I cannot find the exact logic.

Since this seems to be a task designed as coding practice, I'll only guide. No code for you, sir, though the logic and the code aren't that far separated.
You will need to loop through each character of the string, and determine whether or not the character is the delimiter (comma or semicolon, for instance). If not, add it to the last element of the array you plan to return. If it is the delimiter, create a new empty string as the array's last element to start feeding your characters into.

I'm going to assume that this is homework, so I will only give snippets as hints:
Finding indices of all occurrences of a given substring
Here's an example of using indexOf with the fromIndex parameter to find all occurrences of a substring within a larger string:
String text = "012ab567ab0123ab";
// finding all occurrences forward: Method #1
for (int i = text.indexOf("ab"); i != -1; i = text.indexOf("ab", i+1)) {
System.out.println(i);
} // prints "3", "8", "14"
// finding all occurrences forward: Method #2
for (int i = -1; (i = text.indexOf("ab", i+1)) != -1; ) {
System.out.println(i);
} // prints "3", "8", "14"
String API links
int indexOf(String, int fromIndex)
Returns the index within this string of the first occurrence of the specified substring, starting at the specified index. If no such occurrence exists, -1 is returned.
Related questions
Searching for one string in another string
Extracting substrings at given indices out of a string
This snippet extracts substring at given indices out of a string and puts them into a List<String>:
String text = "0123456789abcdefghij";
List<String> parts = new ArrayList<String>();
parts.add(text.substring(0, 5));
parts.add(text.substring(3, 7));
parts.add(text.substring(9, 13));
parts.add(text.substring(18, 20));
System.out.println(parts); // prints "[01234, 3456, 9abc, ij]"
String[] partsArray = parts.toArray(new String[0]);
Some key ideas:
Effective Java 2nd Edition, Item 25: Prefer lists to arrays
Works especially nicely if you don't know how many parts there'll be in advance
String API links
String substring(int beginIndex, int endIndex)
Returns a new string that is a substring of this string. The substring begins at the specified beginIndex and extends to the character at index endIndex - 1.
Related questions
Fill array with List data

You do now that most of the java standard libraries are open source
In this case you can start here

Use String tokenizer to split strings in Java without split:
import java.util.StringTokenizer;
public class tt {
public static void main(String a[]){
String s = "012ab567ab0123ab";
String delims = "ab ";
StringTokenizer st = new StringTokenizer(s, delims);
System.out.println("No of Token = " + st.countTokens());
while (st.hasMoreTokens()) {
System.out.println(st.nextToken());
}
}
}

This is the right answer
import java.util.StringTokenizer;
public class tt {
public static void main(String a[]){
String s = "012ab567ab0123ab";
String delims = "ab ";
StringTokenizer st = new StringTokenizer(s, delims);
System.out.println("No of Token = " + st.countTokens());
while (st.hasMoreTokens())
{
System.out.println(st.nextToken());
}
}
}

/**
* My method split without javas split.
* Return array with words after mySplit from two texts;
* Uses trim.
*/
public class NoJavaSplit {
public static void main(String[] args) {
String text1 = "Some text for example ";
String text2 = " Second sentences ";
System.out.println(Arrays.toString(mySplit(text1, text2)));
}
private static String [] mySplit(String text1, String text2) {
text1 = text1.trim() + " " + text2.trim() + " ";
char n = ' ';
int massValue = 0;
for (int i = 0; i < text1.length(); i++) {
if (text1.charAt(i) == n) {
massValue++;
}
}
String[] splitArray = new String[massValue];
for (int i = 0; i < splitArray.length; ) {
for (int j = 0; j < text1.length(); j++) {
if (text1.charAt(j) == n) {
splitArray[i] = text1.substring(0, j);
text1 = text1.substring(j + 1, text1.length());
j = 0;
i++;
}
}
return splitArray;
}
return null;
}
}

you can try, the way i did `{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
for(int i = 0; i <str.length();i++) {
if(str.charAt(i)==' ') { // whenever it found space it'll create separate words from string
System.out.println();
continue;
}
System.out.print(str.charAt(i));
}
sc.close();
}`

The logic is: go through the whole string starting from first character and whenever you find a space copy the last part to a new string.. not that hard?

The way to go is to define the function you need first. In this case, it would probably be:
String[] split(String s, String separator)
The return type doesn't have to be an array. It can also be a list:
List<String> split(String s, String separator)
The code would then be roughly as follows:
start at the beginning
find the next occurence of the delimiter
the substring between the end of the previous delimiter and the start of the current delimiter is added to the result
continue with step 2 until you have reached the end of the string
There are many fine points that you need to consider:
What happens if the string starts or ends with the delimiter?
What if multiple delimiters appear next to each other?
What should be the result of splitting the empty string? (1 empty field or 0 fields)

You can do it using Java standard libraries.
Say the delimiter is : and
String s = "Harry:Potter"
int a = s.find(delimiter);
and then add
s.substring(start, a)
to a new String array.
Keep doing this till your start < string length
Should be enough I guess.

public class MySplit {
public static String[] mySplit(String text,String delemeter){
java.util.List<String> parts = new java.util.ArrayList<String>();
text+=delemeter;
for (int i = text.indexOf(delemeter), j=0; i != -1;) {
parts.add(text.substring(j,i));
j=i+delemeter.length();
i = text.indexOf(delemeter,j);
}
return parts.toArray(new String[0]);
}
public static void main(String[] args) {
String str="012ab567ab0123ab";
String delemeter="ab";
String result[]=mySplit(str,delemeter);
for(String s:result)
System.out.println(s);
}
}

public class WithoutSpit_method {
public static void main(String arg[])
{
char[]str;
String s="Computer_software_developer_gautam";
String s1[];
for(int i=0;i<s.length()-1;)
{
int lengh=s.indexOf("_",i);
if(lengh==-1)
{
lengh=s.length();
}
System.out.print(" "+s.substring(i,lengh));
i=lengh+1;
}
}
}
Result: Computer software developer gautam

Here is my way of doing with Scanner;
import java.util.Scanner;
public class spilt {
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter the String to be Spilted : ");
String st = input.nextLine();
Scanner str = new Scanner(st);
while (str.hasNext())
{
System.out.println(str.next());
}
}
}
Hope it Helps!!!!!

public class StringWitoutPre {
public static void main(String[] args) {
String str = "md taufique reja";
int len = str.length();
char ch[] = str.toCharArray();
String tmp = " ";
boolean flag = false;
for (int i = 0; i < str.length(); i++) {
if (ch[i] != ' ') {
tmp = tmp + ch[i];
flag = false;
} else {
flag = true;
}
if (flag || i == len - 1) {
System.out.println(tmp);
tmp = " ";
}
}
}
}

In Java8 we can use Pattern and get the things done in more easy way. Here is the code.
package com.company;
import java.util.regex.Pattern;
public class umeshtest {
public static void main(String a[]) {
String ss = "I'm Testing and testing the new feature";
Pattern.compile(" ").splitAsStream(ss).forEach(s -> System.out.println(s));
}
}

static void splitString(String s, int index) {
char[] firstPart = new char[index];
char[] secondPart = new char[s.length() - index];
int j = 0;
for (int i = 0; i < s.length(); i++) {
if (i < index) {
firstPart[i] = s.charAt(i);
} else {
secondPart[j] = s.charAt(i);
if (j < s.length()-index) {
j++;
}
}
}
System.out.println(firstPart);
System.out.println(secondPart);
}

import java.util.Scanner;
public class Split {
static Scanner in = new Scanner(System.in);
static void printArray(String[] array){
for (int i = 0; i < array.length; i++) {
if(i!=array.length-1)
System.out.print(array[i]+",");
else
System.out.println(array[i]);
}
}
static String delimeterTrim(String str){
char ch = str.charAt(str.length()-1);
if(ch=='.'||ch=='!'||ch==';'){
str = str.substring(0,str.length()-1);
}
return str;
}
private static String [] mySplit(String text, char reg, boolean delimiterTrim) {
if(delimiterTrim){
text = delimeterTrim(text);
}
text = text.trim() + " ";
int massValue = 0;
for (int i = 0; i < text.length(); i++) {
if (text.charAt(i) == reg) {
massValue++;
}
}
String[] splitArray = new String[massValue];
for (int i = 0; i < splitArray.length; ) {
for (int j = 0; j < text.length(); j++) {
if (text.charAt(j) == reg) {
splitArray[i] = text.substring(0, j);
text = text.substring(j + 1, text.length());
j = 0;
i++;
}
}
return splitArray;
}
return null;
}
public static void main(String[] args) {
System.out.println("Enter the sentence :");
String text = in.nextLine();
//System.out.println("Enter the regex character :");
//char regex = in.next().charAt(0);
System.out.println("Do you want to trim the delimeter ?");
String delch = in.next();
boolean ch = false;
if(delch.equalsIgnoreCase("yes")){
ch = true;
}
System.out.println("Output String array is : ");
printArray(mySplit(text,' ',ch));
}
}

Split a string without using split()
static String[] splitAString(String abc, char splitWith){
char[] ch=abc.toCharArray();
String temp="";
int j=0,length=0,size=0;
for(int i=0;i<abc.length();i++){
if(splitWith==abc.charAt(i)){
size++;
}
}
String[] arr=new String[size+1];
for(int i=0;i<ch.length;i++){
if(length>j){
j++;
temp="";
}
if(splitWith==ch[i]){
length++;
}else{
temp +=Character.toString(ch[i]);
}
arr[j]=temp;
}
return arr;
}
public static void main(String[] args) {
String[] arr=splitAString("abc-efg-ijk", '-');
for(int i=0;i<arr.length;i++){
System.out.println(arr[i]);
}
}
}

You cant split with out using split(). Your only other option is to get the strings char indexes and and get sub strings.