Removing Whitespaces Using Regex in Java
In this section, we will write Java programs that remove whitespaces using Regex. To remove whitespaces in Java, use a regular expression pattern to match and replace all whitespace characters in a string. Whitespace characters consist of spaces, tabs, and newline characters.
Example:
Input: String input = "Tutorial and Example"
Output: Original String: Tutorial and Example /p
Result: TutorialandExample
Methods for Removing whitespaces with Regex:
There are several approaches to whitespace removal using regex; here are some methods:
- Using replaceAll() Method
- Using Pattern and Matcher classes
- Using String.split and String.join with Regex
- Using Regular Expression and Java 8 Streams
- Using Scanner and Regular Expression
Using replaceAll() Method
The Java code implements a removeWhitespace() method that uses the replaceAll() method and a regex (\\s+) to remove all whitespaces from the input string.
FileName: RemoveSpaces.java
public class RemoveSpaces {
// Method to remove whitespaces using replaceAll
public static String removeWhitespace(String input) {
// Use regex \s+ to match and replace whitespaces
return input.replaceAll("\\s+", "");
}
public static void main(String[] args) {
// Example usage
String input = "Tutorial and Example";
System.out.println("Original String: " + input);
String result = removeWhitespace(input);
System.out.println("Result: " + result);
}
}
Output:
Original String: Tutorial and Example
Result: TutorialandExample
Complexity Analysis: The code has a time complexity of O(n), where n is the input string's length because it iterates through each character once. The space complexity is O(n), as a new string is produced to store the result.
Using Pattern and Matcher classes
The Java code implements a removeWhitespace() method that uses the Pattern and Matcher classes to generate a regex pattern (\\s+) and replace all whitespaces in the input string.
FileName: RemoveSpaces1.java
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class RemoveSpaces1 {
// Method to remove whitespaces using Pattern and Matcher
public static String removeWhitespace(String input) {
// Create a Pattern with regex \s+
Pattern pattern = Pattern.compile("\\s+");
// Use Matcher to find and replace whitespaces
Matcher matcher = pattern.matcher(input);
return matcher.replaceAll("");
}
public static void main(String[] args) {
String input = "Tutorial and Example";
System.out.println("Original String: " + input);
String result = removeWhitespace(input);
System.out.println("Result: " + result);
}
}
Output:
Original String: Tutorial and Example
Result: TutorialandExample
Complexity Analysis: The code's time and space complexity is O(n), where n represents the length of the input string.
Using String.split and String.join
The Java code specifies a removeWhitespace() method that uses String.split("\\s+") to split the input string into parts depending on one or more whitespaces, and then String.join("") to concatenate the portions without whitespaces.
FileName: RemoveSpaces2.java
import java.util.Arrays;
public class RemoveSpaces2 {
// Method to remove whitespaces using String.split and String.join with regex
public static String removeWhitespace(String input) {
// Split the input using regex \s+ and join the parts
return String.join("", input.split("\\s+"));
}
public static void main(String[] args) {
String input = "Tutorial and Example";
System.out.println("Original String: " + input);
String result = removeWhitespace(input);
System.out.println("Result: " + result);
}
}
Output:
Original String: Tutorial and Example
Result: TutorialandExample
Complexity Analysis: The code's time and space complexity is O(n), where n represents the length of the input string.
Using Regular Expression and Java 8 Streams
The Java code defines a removeWhitespace() function that uses Java 8 Streams and Patterns.compile("\\s+").splitAsStream(input) splits an input string into non-whitespace tokens and Collectors.joining() to concatenate them.
FileName: RemoveSpaces3.java
import java.util.regex.Pattern;
import java.util.stream.Collectors;
public class RemoveSpaces3 {
// Method to remove whitespaces using regular expression and Java 8 Streams
public static String removeWhitespace(String input) {
return Pattern.compile("\\s+")
.splitAsStream(input)
.collect(Collectors.joining());
}
public static void main(String[] args) {
String input = "Tutorial and Example";
System.out.println("Original String: " + input);
String result = removeWhitespace(input);
System.out.println("Result: " + result);
}
}
Output:
Original String: Tutorial and Example
Result: TutorialandExample
Complexity Analysis: The removeWhitespace() function has a time complexity of O(n), where n is the length of the input string because the splitAsStream() action only iterates through the characters once. The space complexity is O(n) because the split operation generates a stream of substrings, which are then joined into a new string.
Using Scanner and Regular Expression
The code uses a Scanner and regular expressions to remove excessive spaces from an input string. The removeWhitespace() method uses a Scanner with the regular expression \\s+ as a delimiter to break the input string into tokens separated by one or more whitespace characters. The collected tokens are then concatenated with a StringBuilder.
FileName: RemoveSpaces4.java
import java.util.Scanner;
public class RemoveSpaces4 {
// Method to remove whitespaces using Scanner and regular expression
public static String removeWhitespace(String input) {
try (Scanner scanner = new Scanner(input)) {
// Use \\s+ regex as a delimiter and concatenate the parts
return scanner.useDelimiter("\\s+").tokens().collect(StringBuilder::new, StringBuilder::append, StringBuilder::append).toString();
}
}
public static void main(String[] args) {
String input = "Tutorial and Example";
System.out.println("Original String: " + input);
String result = removeWhitespace(input);
System.out.println("Result: " + result);
}
}
Output:
Original String: Tutorial and Example
Result: TutorialandExample
Complexity Analysis: The time complexity of the code is O(n), where n is the length of the input string because each character is processed once. The space complexity is also O(n), as the Scanner tokens are collected into a StringBuilder, and the necessary space is proportional to the length of the input string.