Strings
Strings
JUnit
recursion
reverse
package com.company;
import java.util.Scanner;
public class Main {
public static String reverse(String str){
String reversed = "";
for (int i = 0; i < str.length(); i++)
reversed = str.charAt(i) + reversed;
return reversed;
}
public static void main(String[] args) {
String str = "hello";
System.out.println(reverse(str));
}
}ollehJUnit
import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
import static com.company.Main.reverse;
public class MainTests {
@Test
void TestReverse() {
Assertions.assertEquals("",reverse(""));
Assertions.assertEquals("olleh",reverse("hello"));
Assertions.assertEquals("dlrow olleh",reverse("hello world"));
}
}passedversion 2
public static String reverse(String str){
String reversed = "";
for (char c : str.toCharArray())
reversed = c + reversed;
return reversed;
}JUnit on array
Lets add this function
public class Main {
//added this function
public static String[] reverse_strings(String[] strings){
String[] r_strings = new String[strings.length];
for (int i = 0; i < strings.length; i++) {
r_strings[i] = reverse(strings[i]);
}
return r_strings;
}
//same as before
public static String reverse(String str){
String reversed = "";
for (int i = 0; i < str.length(); i++)
reversed = str.charAt(i) + reversed;
return reversed;
}
public static void main(String[] args) {
String[] strings = {"hello", "world"};
reverse_strings(strings);
}
}and add some more tests
import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
import static com.company.Main.reverse;
import static com.company.Main.reverse_strings;
public class MainTests {
@Test
void TestReverse() {
Assertions.assertEquals("dlrow olleh",reverse("hello world"));
}
@Test
void TestReverseArray() {
String[] strings = {"hello", "world"};
Assertions.assertArrayEquals(new String[]{"olleh", "dlrow"},
reverse_strings(new String[]{"hello", "world"}));
}
}here we used assertArrayEquals
Exercise - Permutations
public class Main {
public static int occurances_of_char(String str, char c) {
int count=0;
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == c)
++count;
}
return count;
}
public static boolean permutation(String str1, String str2){
if (str1.length()!=str2.length())
return false;
for (int i = 0; i < str1.length(); i++) {
char curr_char = str1.charAt(i);
if(occurances_of_char(str1, curr_char) != occurances_of_char(str2, curr_char))
return false;
}
return true;
}
public static void main(String[] args) {
String str = "hello";
System.out.println(permutation(str, "lleho"));
System.out.println(permutation(str, "llehol"));
System.out.println(permutation(str, ""));
}
}true
false
falseversion 2
since the version above is 0(n^2) it isn't very efficient, it would be better to use a bucket sort type of approach and that would give of 2*O(n)
Exercise: Remove words from String
Write a function that receives a string and 2 others string to be removed
remove2Strings("zzzabzzzabzzzabc", "ab", "c") -> returns -> "zzzzzzzzz"solution
public class Main {
public static void main(String[] args) {
String str = remove2Strings("zzzabzzzabzzzabc", "ab", "c");
System.out.println(str);
str = removeManyStrings("zzzabzzzbczzze", "ab", "bc", "e");
System.out.println(str);
}
public static String remove2Strings(String str, String getRidOf1, String getRidOf2) {
String res = str.replace(getRidOf1, "").replace(getRidOf2, "");
return res;
}
//... - means many args
public static String removeManyStrings(String str, String... args) {
String res = str;
for(int i = 0; i < args.length; i++){
res = res.replace(args[i], "");
}
return res;
}
}zzzzzzzzz
zzzzzzzzz
String... args- means there are an unknow amount of Strings (an array of Strings)
Exercise: Calculate the sum of the numbers appear in a given string
string 12 with120 20numbers -> returns 152Solution
public class Main {
public static int sumOfTheNumbers(String str) {
int len = str.length();
int sum = 0;
String temp = "";
for (int i = 0; i < len; i++) {
//Character.isDigit() - Determines if the specified character is a digit.
if (Character.isDigit(str.charAt(i))) {
//if next char is also a digit, append
if (i < len-1 && Character.isDigit(str.charAt(i+1))) {
temp += str.charAt(i);
} else { //if next char isn't a digit (but this char is) or end of the string
temp += str.charAt(i);
sum += Integer.parseInt(temp);
temp = "";
}
}
}
return sum;
}
public static void main (String[] args) {
String str1 = "this 12 is120 a string 20with numbers";
System.out.println("The given string is: "+str1);
System.out.println("The sum of numbers in the string is: "+ sumOfTheNumbers(str1));
}
}The given string is: this 12 is120 a string 20with numbers
The sum of numbers in the string is: 152Exercise: Convert Int to Roman
I can be placed before V or X, represents subtract one, so IV (5-1) = 4 and 9 is IX (10-1)=9.
X can be placed before L or C represents subtract ten, so XL (50-10) = 40 and XC (100-10)=90.
C placed before D or M represents subtract hundred, so CD (500-100)=400 and CM (1000-100)=900.
for example
4 -> returns -> IV
25 -> returns -> XXV
400 -> returns -> CMSolution
public class Main {
public static String intToRoman(int num) {
System.out.println("Integer: " + num);
int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
String[] romanLiterals = {"M", "CM", "D", "CD", "C", "XC","L","XL","X","IX","V","IV","I"};
//StringBuilder represents a mutable sequence of characters.
//String Class creates an immutable sequence of characters,
StringBuilder roman = new StringBuilder();
for(int i=0;i<values.length;i++) {
while(num >= values[i]) {
num -= values[i];
roman.append(romanLiterals[i]);
}
}
return roman.toString();
}
public static void main(String[] args) {
System.out.println(intToRoman(25) + "\n");
System.out.println(intToRoman(1023)+ "\n");
System.out.println(intToRoman(511) + "\n");
}
}Integer: 25
XXV
Integer: 1023
MXXIII
Integer: 511
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