Java Foundations: Strings and Text Processing

Java Foundations
Strings and Text
Processing

Your Course
Instructors
Svetlin Nakov
George Georgiev

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Text Processing
Manipulating Text
6

Table of Contents
1. What Is a String?
2. Manipulating Strings
3. Building and Modifying Strings
 Using StringBuilder Class
 Why Concatenation Is a Slow
Operation?
7

 Strings are sequences of characters (texts)
 The string data type in Java
 Declared by the String
 Strings are enclosed in double quotes:
What Is a String?
9
String text = "Hello, Java";

 Strings are immutable (read-only)
sequences of characters
 Accessible by index (read-only)
 Strings use Unicode
(can use most alphabets, e.g. Arabic)
Strings Are Immutable
10
String str = "Hello, Java";
char ch = str.charAt(2); // l
String greeting = "你好"; // (lí-hó) Taiwanese

 Initializing from a string literal:
 Reading a string from the console:
 Converting a string from and to a char array:
Initializing a String
11
String str = "Hello, Java";
String name = sc.nextLine();
System.out.println("Hi, " + name);
String str = new String(new char[] {'s', 't', 'r'});
char[] charArr = str.toCharArray();
// ['s', 't', 'r']

Concatenating
 Use the + or the += operators
 Use the concat() method
String greet = "Hello, ";
String name = "John";
String result = greet.concat(name);
System.out.println(result); // "Hello, John"
String text = "Hello, ";
text += "John"; // "Hello, John"
String text = "Hello" + ", " + "world!";
// "Hello, world!"
13

Joining Strings
 String.join("", …) concatenates strings
 Or an array/list of strings
 Useful for repeating a string
14
String t = String.join("", "con", "ca", "ten", "ate");
// "concatenate"
String s = "abc";
String[] arr = new String[3];
for (int i = 0; i < arr.length; i++) { arr[i] = s; }
String repeated = String.join("", arr); // "abcabcabc"

 Read an array from strings
 Repeat each word n times, where n is the length of the word
Problem: Repeat Strings
15
hi abc add hihiabcabcabcaddaddadd
work workworkworkwork
ball ballballballball

String[] words = sc.nextLine().split(" ");
List<String> result = new ArrayList<>();
for (String word : words) {
result.add(repeat(word, word.length()));
}
System.out.println(String.join("", result));
Solution: Repeat Strings (1)
16

static String repeat(String s, int repeatCount) {
String[] repeatArr = new String[repeatCount];
for (int i = 0; i < repeatCount; i++) {
repeatArr[i] = s;
}
return String.join("", repeatArr);
}
Solution: Repeat Strings (2)
17

Substring
 substring(int startIndex, int endIndex)
 substring(int startIndex)
18
String text = "My name is John";
String extractWord = text.substring(11);
System.out.println(extractWord); // John
String card = "10C";
String power = card.substring(0, 2);
System.out.println(power); // 10

Searching (1)
 indexOf() - returns the first match index or -1
 lastIndexOf() - finds the last occurrence
19
String fruits = "banana, apple, kiwi, banana, apple";
System.out.println(fruits.lastIndexOf("banana")); // 21
System.out.println(fruits.lastIndexOf("orange")); // -1
String fruits = "banana, apple, kiwi, banana, apple";
System.out.println(fruits.indexOf("banana")); // 0
System.out.println(fruits.indexOf("orange")); // -1

Searching (2)
 contains() - checks whether one string
contains another
20
String text = "I love fruits.";
System.out.println(text.contains("fruits"));
// true
System.out.println(text.contains("banana"));
// false

 You are given a remove word and a text
 Remove all substrings that are equal to the remove word
Problem: Substring
21
ice
kicegiceiceb
kgb
abc
tabctqw
ttqw
key
keytextkey
text
word
wordawordbwordc
abc

String key = sc.nextLine();
String text = sc.nextLine();
int index = text.indexOf(key);
while (index != -1) {
text = text.replace(key, "");
index = text.indexOf(key);
}
System.out.println(text);
Solution: Substring
22

Splitting
 Split a string by given pattern
 Split by multiple separators
23
String text = "Hello, I am John.";
String[] words = text.split("[, .]+");
// "Hello", "I", "am", "John"
String text = "Hello, john@softuni.org, you have been
using john@softuni.org in your registration";
String[] words = text.split(", ");
// words[]: "Hello","john@softuni.org","you have been…"

Replacing
 replace(match, replacement) - replaces all
occurrences
 The result is a new string (strings are immutable)
24
String text = "Hello, john@softuni.org, you have been
using john@softuni.org in your registration.";
String replacedText = text
.replace("john@softuni.org", "john@softuni.com");
System.out.println(replacedText);
// Hello, john@softuni.com, you have been using
john@softuni.com in your registration.

 You are given a string of banned words and a text
 Replace all banned words in the text with asterisks (*)
Problem: Text Filter
25
Linux, Windows
It is not Linux, it is GNU/Linux. Linux is merely
the kernel, while GNU adds the functionality...
It is not *****, it is GNU/*****. ***** is merely
the kernel, while GNU adds the functionality...

String[] banWords = sc.nextLine().split(", ");
String text = sc.nextLine();
for (String banWord : banWords) {
if (text.contains(banWord)) {
String replacement = repeatStr("*",
banWord.length());
text = text.replace(banWord, replacement);
}
}
Solution: Text Filter (1)
26
contains(…) checks if string
contains another string
replace() a word with a sequence
of asterisks of the same length

private static String repeatStr(String str, int length) {
String replacement = "";
for (int i = 0; i < length; i++) {
replacement += str;
}
return replacement;
}
Solution: Text Filter (2)
27

Building and Modifying Strings
Using the StringBuilder Class

 StringBuilder keeps a buffer space, allocated in advance
 Do not allocate memory for
most operations  performance
StringBuilder: How It Works?
H e l l o , J a v a
StringBuilder:
length() = 10
capacity() = 16
Capacity
used buffer
(Length)
unused
buffer
30

Using StringBuilder Class
 Use the StringBuilder to build/modify strings
31
StringBuilder sb = new StringBuilder();
sb.append("Hello, ");
sb.append("John! ");
sb.append("I sent you an email.");
System.out.println(sb.toString());
// Hello, John! I sent you an email.

Concatenation vs. StringBuilder (1)
 Concatenating strings is a slow operation
because each iteration creates a new string
32
Tue Jul 10 13:57:20 EEST 2021
Tue Jul 10 13:58:07 EEST 2021
System.out.println(new Date());
String text = "";
for (int i = 0; i < 1000000; i++)
text += "a";

Concatenation vs. StringBuilder (2)
 Using StringBuilder
33
Tue Jul 10 14:51:31 EEST 2021
Tue Jul 10 14:51:31 EEST 2021
StringBuilder text = new
StringBuilder();
for (int i = 0; i < 1000000; i++)
text.append("a");

StringBuilder Methods (1)
 append() - appends the string representation
of the argument
 length() - holds the length of the string in the buffer
 setLength(0) - removes all characters
34
sb.append("Hello Peter, how are you?");
System.out.println(sb.length()); // 25

 charAt(int index) - returns char on index
 insert(int index, String str) –
inserts a string at the specified character position
35
System.out.println(sb.charAt(1)); // e
sb.insert(11, " Ivanov");
System.out.println(sb);
// Hello Peter Ivanov, how are you?

 replace(int startIndex, int endIndex,
String str) - replaces the chars in a substring
 toString() - converts the value of this instance
to a String
36
String text = sb.toString();
// Hello George, how are you?
sb.replace(6, 11, "George");

 …
 …
 …
Summary
38
 Strings are immutable sequences of
Unicode characters
 String processing methods
 concat(), indexOf(), contains(),
substring(), split(), replace(), …
 StringBuilder efficiently builds/modifies
strings

 …
 …
 …
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Java Foundations: Strings and Text Processing

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