How can I check if a single character appears in a string?

Question

In Java is there a way to check the condition:

"Does this single character appear at all in string x"

without using a loop?

Answer 1

You can use string.indexOf('a').

If the char a is present in string :

it returns the the index of the first occurrence of the character in the character sequence represented by this object, or -1 if the character does not occur.

Answer 2

• String.contains() which checks if the string contains a specified sequence of char values
• String.indexOf() which returns the index within the string of the first occurence of the specified character or substring (there are 4 variations of this method)

Answer 3

I'm not sure what the original poster is asking exactly. Since indexOf(...) and contains(...) both probably use loops internally, perhaps he's looking to see if this is possible at all without a loop? I can think of two ways off hand, one would of course be recurrsion:

public boolean containsChar(String s, char search) {
    if (s.length() == 0)
        return false;
    else
        return s.charAt(0) == search || containsChar(s.substring(1), search);
}

The other is far less elegant, but completeness...:

/**
 * Works for strings of up to 5 characters
 */
public boolean containsChar(String s, char search) {
    if (s.length() > 5) throw IllegalArgumentException();

    try {
        if (s.charAt(0) == search) return true;
        if (s.charAt(1) == search) return true;
        if (s.charAt(2) == search) return true;
        if (s.charAt(3) == search) return true;
        if (s.charAt(4) == search) return true;
    } catch (IndexOutOfBoundsException e) {
        // this should never happen...
        return false;
    }
    return false;
}

The number of lines grow as you need to support longer and longer strings of course. But there are no loops/recurrsions at all. You can even remove the length check if you're concerned that that length() uses a loop.

Answer 4

You can use 2 methods from the String class.

• String.contains() which checks if the string contains a specified sequence of char values
• String.indexOf() which returns the index within the string of the first occurence of the specified character or substring or returns -1 if the character is not found (there are 4 variations of this method)

Method 1:

String myString = "foobar";
if (myString.contains("x") {
    // Do something.
}

Method 2:

String myString = "foobar";
if (myString.indexOf("x") >= 0 {
    // Do something.
}

Links by: Zach Scrivena

Answer 5

String temp = "abcdefghi";
if(temp.indexOf("b")!=-1)
{
   System.out.println("there is 'b' in temp string");
}
else
{
   System.out.println("there is no 'b' in temp string");
}

Answer 6

If you need to check the same string often you can calculate the character occurrences up-front. This is an implementation that uses a bit array contained into a long array:

public class FastCharacterInStringChecker implements Serializable {
private static final long serialVersionUID = 1L;

private final long[] l = new long[1024]; // 65536 / 64 = 1024

public FastCharacterInStringChecker(final String string) {
    for (final char c: string.toCharArray()) {
        final int index = c >> 6;
        final int value = c - (index << 6);
        l[index] |= 1L << value;
    }
}

public boolean contains(final char c) {
    final int index = c >> 6; // c / 64
    final int value = c - (index << 6); // c - (index * 64)
    return (l[index] & (1L << value)) != 0;
}}

Answer 7

To check if something does not exist in a string, you at least need to look at each character in a string. So even if you don't explicitly use a loop, it'll have the same efficiency. That being said, you can try using str.contains(""+char).

Answer 8

Is the below what you were looking for?

int index = string.indexOf(character);
return index != -1;

Answer 9

Yes, using the indexOf() method on the string class. See the API documentation for this method

Answer 10

String.contains(String) or String.indexOf(String) - suggested

"abc".contains("Z"); // false - correct
"zzzz".contains("Z"); // false - correct
"Z".contains("Z"); // true - correct
"and".contains(""); // true - correct
"and".contains(""); // false - correct
"and".indexOf(""); // 0 - correct
"and".indexOf(""); // -1 - correct

String.indexOf(int) and carefully considered String.indexOf(char) with char to int widening

"and".indexOf("".charAt(0)); // 0 though incorrect usage has correct output due to portion of correct data
"and".indexOf("".charAt(0)); // 0 -- incorrect usage and ambiguous result
"and".indexOf("".codePointAt(0)); // -1 -- correct usage and correct output

The discussions around character is ambiguous in Java world

can the value of char or Character considered as single character?

No. In the context of unicode characters, char or Character can sometimes be part of a single character and should not be treated as a complete single character logically.

if not, what should be considered as single character (logically)?

Any system supporting character encodings for Unicode characters should consider unicode's codepoint as single character.

So Java should do that very clear & loud rather than exposing too much of internal implementation details to users.

String class is bad at abstraction (though it requires confusingly good amount of understanding of its encapsulations to understand the abstraction and hence an anti-pattern).

How is it different from general char usage?

char can be only be mapped to a character in Basic Multilingual Plane.

Only codePoint - int can cover the complete range of Unicode characters.

Why is this difference?

char is internally treated as 16-bit unsigned value and could not represent all the unicode characters using UTF-16 internal representation using only 2-bytes. Sometimes, values in a 16-bit range have to be combined with another 16-bit value to correctly define character.

Without getting too verbose, the usage of indexOf, charAt, length and such methods should be more explicit. Sincerely hoping Java will add new UnicodeString and UnicodeCharacter classes with clearly defined abstractions.

Reason to prefer contains and not indexOf(int)

1. Practically there are many code flows that treat a logical character as char in java.
2. In Unicode context, char is not sufficient
3. Though the indexOf takes in an int, char to int conversion masks this from the user and user might do something like str.indexOf(someotherstr.charAt(0))(unless the user is aware of the exact context)
4. So, treating everything as CharSequence (aka String) is better

    public static void main(String[] args) {
        System.out.println("and".indexOf("".charAt(0))); // 0 though incorrect usage has correct output due to portion of correct data
        System.out.println("and".indexOf("".charAt(0))); // 0 -- incorrect usage and ambiguous result
        System.out.println("and".indexOf("".codePointAt(0))); // -1 -- correct usage and correct output
        System.out.println("and".contains("")); // true - correct
        System.out.println("and".contains("")); // false - correct
    }

Semantics

1. char can handle most of the practical use cases. Still its better to use codepoints within programming environment for future extensibility.
2. codepoint should handle nearly all of the technical use cases around encodings.
3. Still, Grapheme Clusters falls out of the scope of codepoint level of abstraction.
4. Storage layers can choose char interface if ints are too costly(doubled). Unless storage cost is the only metric, its still better to use codepoint. Also, its better to treat storage as byte and delegate semantics to business logic built around storage.
5. Semantics can be abstracted at multiple levels. codepoint should become lowest level of interface and other semantics can be built around codepoint in runtime environment.

Answer 11

package com;
public class _index {

    public static void main(String[] args) {
        String s1="be proud to be an indian";
        char ch=s1.charAt(s1.indexOf('e'));
        int count = 0; 
        for(int i=0;i<s1.length();i++) {
            if(s1.charAt(i)=='e'){
                System.out.println("number of E:=="+ch);
                count++;
            }
        }
        System.out.println("Total count of E:=="+count);
    }
}

Answer 12

static String removeOccurences(String a, String b)
{
    StringBuilder s2 = new StringBuilder(a);

    for(int i=0;i<b.length();i++){
        char ch = b.charAt(i);  
        System.out.println(ch+"  first index"+a.indexOf(ch));

        int lastind = a.lastIndexOf(ch);

    for(int k=new String(s2).indexOf(ch);k > 0;k=new String(s2).indexOf(ch)){
            if(s2.charAt(k) == ch){
                s2.deleteCharAt(k);
        System.out.println("val of s2 :             "+s2.toString());
            }
        }
      }

    System.out.println(s1.toString());

    return (s1.toString());
}

Answer 13

you can use this code. It will check the char is present or not. If it is present then the return value is >= 0 otherwise it's -1. Here I am printing alphabets that is not present in the input.

import java.util.Scanner;

public class Test {

public static void letters()
{
    System.out.println("Enter input char");
    Scanner sc = new Scanner(System.in);
    String input = sc.next();
    System.out.println("Output : ");
    for (char alphabet = 'A'; alphabet <= 'Z'; alphabet++) {
            if(input.toUpperCase().indexOf(alphabet) < 0) 
                System.out.print(alphabet + " ");
    }
}
public static void main(String[] args) {
    letters();
}

}

//Ouput Example
Enter input char
nandu
Output : 
B C E F G H I J K L M O P Q R S T V W X Y Z

Answer 14

If you see the source code of indexOf in JAVA:

public int indexOf(int ch, int fromIndex) {

        final int max = value.length;

        if (fromIndex < 0) {

            fromIndex = 0;

        } else if (fromIndex >= max) {

            // Note: fromIndex might be near -1>>>1.

            return -1;

        }


        if (ch < Character.MIN_SUPPLEMENTARY_CODE_POINT) {

            // handle most cases here (ch is a BMP code point or a

            // negative value (invalid code point))

            final char[] value = this.value;

            for (int i = fromIndex; i < max; i++) {

                if (value[i] == ch) {

                    return i;

                }

            }

            return -1;

        } else {

            return indexOfSupplementary(ch, fromIndex);

        }

    }

you can see it uses a for loop for finding a character. Note that each indexOf you may use in your code, is equal to one loop.

So, it is unavoidable to use loop for a single character.

However, if you want to find a special string with more different forms, use useful libraries such as util.regex, it deploys stronger algorithm to match a character or a string pattern with Regular Expressions. For example to find an email in a string:

String regex = "^(.+)@(.+)$";
 
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(email);

If you don't like to use regex, just use a loop and charAt and try to cover all cases in one loop.

Be careful recursive methods has more overhead than loop, so it's not recommended.

Answer 15

how about one uses this ;

let text = "Hello world, welcome to the universe.";
let result = text.includes("world");
console.log(result) ....// true

the result will be a true or false

this always works for me

Answer 16

You won't be able to check if char appears at all in some string without atleast going over the string once using loop / recursion ( the built-in methods like indexOf also use a loop )

If the no. of times you look up if a char is in string x is more way more than the length of the string than I would recommend using a Set data structure as that would be more efficient than simply using indexOf

String s = "abc";

// Build a set so we can check if character exists in constant time O(1)
Set<Character> set = new HashSet<>();
int len = s.length();
for(int i = 0; i < len; i++) set.add(s.charAt(i));

// Now we can check without the need of a loop
// contains method of set doesn't use a loop unlike string's contains method
set.contains('a') // true
set.contains('z') // false

Using set you will be able to check if character exists in a string in constant time O(1) but you will also use additional memory ( Space complexity will be O(n) ).