How to check if string contains only digits in Java

Given string str, the task is to write a Java program to check whether a string contains only digits or not. If so, then print true, otherwise false. 

Examples:

Input: str = “1234” 
Output: true 
Explanation:
The given string contains only digits so that output is true.

Input: str = “GeeksforGeeks2020” 
Output: false 
Explanation:
The given string contains alphabet character and digits so that output is false.  

Solutions 

1. Using Traversal

The idea is to traverse each character in the string and check if the character of the string contains only digits from 0 to 9. If all the character of the string contains only digits then return true, otherwise, return false. Below is the implementation of the above approach: 

Java

// Java program for the above approach
// contains only digits
class GFG {

    // Function to check if a string
    // contains only digits
    public static boolean
    onlyDigits(String str, int n)
    {
        // Traverse the string from
        // start to end
        for (int i = 0; i < n; i++) {

            // Check if character is
            // not a digit between 0-9
            // then return false
            if (str.charAt(i) < '0'
                || str.charAt(i) > '9') {
                return false;
            }
        }
          // If we reach here, that means
          // all characters were digits.
        return true;
    }

    // Driver Code
    public static void main(String args[])
    {
        // Given string str
        String str = "1a234";
        int len = str.length();

        // Function Call
        System.out.println(onlyDigits(str, len));
    }
}

Output

false

• Time Complexity: O(N), where N is the length of the given string. 
• Auxiliary Space: O(1)  

2. Using Character.isDigit(char ch)

The idea is to iterate over each character of the string and check whether the specified character is a digit or not using Character.isDigit(char ch). If the character is a digit then return true, else return false. Below is the implementation of the above approach:  

Java

// Java program to check if a string
// contains only digits
class GFG {

    // Function to check if a string
    // contains only digits
    public static boolean onlyDigits(String str, int n)
    {

        // Traverse the string from
        // start to end
        for (int i = 0; i < n; i++) {

            // Check if the specified
            // character is a not digit
            // then return false,
            // else return false
            if (!Character.isDigit(str.charAt(i))) {
                return false;
            }
        }
          // If we reach here that means all
          // the characters were digits,
          // so we return true
        return true;
    }

    // Driver Code
    public static void main(String args[])
    {
        // Given string str
        String str = "1234";
        int len = str.length();

        // Function Call
        System.out.println(onlyDigits(str, len));
    }
}

Output

true

• Time Complexity: O(N), where N is the length of the given string. 
• Auxiliary Space: O(1) 

3. Using Regular Expression

• Get the String. 
• Create a Regular Expression to check string contains only digits as mentioned below:  

regex = "[0-9]+";

• Match the given string with Regular Expression. In Java, this can be done by using Pattern.matcher().
• Return true if the string matches with the given regular expression, else return false.

Below is the implementation of the above approach:

Java

// Java program to check if a string
// contains only digits
import java.util.regex.*;
class GFG {

    // Function to validate URL
    // using regular expression
    public static boolean
    onlyDigits(String str)
    {
        // Regex to check string
        // contains only digits
        String regex = "[0-9]+";

        // Compile the ReGex
        Pattern p = Pattern.compile(regex);

        // If the string is empty
        // return false
        if (str == null) {
            return false;
        }

        // Find match between given string
        // and regular expression
        // using Pattern.matcher()
        Matcher m = p.matcher(str);

        // Return if the string
        // matched the ReGex
        return m.matches();
    }

    // Driver Code
    public static void main(String args[])
    {
        // Given string str
        String str = "1234";

        // Function Call
        System.out.println(onlyDigits(str));
    }
}

Output

true

• Time Complexity: O(N), where N is the length of the given string. 
• Auxiliary Space: O(1)  

4. Using contains() method and ArrayList

Java

// Java program for the above approach
// contains only digits
import java.lang.*;
import java.util.*;
class Main{

    // Function to check if a string
    // contains only digits
    public static boolean onlyDigits(String str, int n)
    {
        String num="0123456789";
        ArrayList<Character> numbers = new ArrayList<Character>();
        for(int i=0;i<num.length();i++)
        {
            numbers.add(num.charAt(i));
        }
        // Traverse the string from
        // start to end
        for (int i = 0; i < n; i++) {

            // Check if character is
            // not a digit between 0-9
            // then return false
            if (!numbers.contains(str.charAt(i))) {
                return false;
            }
        }
        // If we reach here, that means
        // all characters were digits.
        return true;
    }

    // Driver Code
    public static void main(String args[])
    {
        // Given string str
        String str = "1a234";
        int len = str.length();

        // Function Call
        System.out.println(onlyDigits(str, len));
    }
}

Output

false