64 lines
2.8 KiB
Plaintext
64 lines
2.8 KiB
Plaintext
import java.util.Scanner;
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public class Lab5 {
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public static void main(String[] args) {
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int C17 = 2430 % 17;
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System.out.println("\n---------------------------------------------------------------------------------------------------------------------");
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System.out.println(" C17 = " + C17 + ", So, the task is: delete all previous occurrences of the last letter of each word of the specified text");
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System.out.println("---------------------------------------------------------------------------------------------------------------------");
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Scanner scanner = new Scanner(System.in);
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String input = null;
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while (input == null || input.isEmpty()) {
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System.out.print("\nEnter a string: ");
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input = scanner.nextLine();
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}
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try {
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StringBuilder sb = new StringBuilder(input);
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String[] words = sb.toString().split("\\s+");
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for (int i = 0; i < words.length; i++) {
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String word = words[i];
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char lastChar = word.charAt(word.length() - 1);
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if (!Character.isLetterOrDigit(lastChar)) {
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// Last character is a punctuation mark
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int lastLetterIndex = -1;
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for (int j = word.length() - 2; j >= 0; j--) {
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if (Character.isLetter(word.charAt(j))) {
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lastLetterIndex = j;
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break;
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}
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}
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if (lastLetterIndex != -1) {
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char lastLetter = word.charAt(lastLetterIndex);
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String newWord = "";
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for (int j = 0; j < word.length() - 1; j++) {
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if (!Character.isLetterOrDigit(word.charAt(j)) || word.charAt(j) != lastLetter) {
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newWord += word.charAt(j);
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}
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}
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newWord += lastChar;
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words[i] = newWord;
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}
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} else {
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// Last character is a letter or a digit
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char lastLetter = lastChar;
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String newWord = "";
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for (int j = 0; j < word.length() - 1; j++) {
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if (!Character.isLetterOrDigit(word.charAt(j)) || word.charAt(j) != lastLetter) {
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newWord += word.charAt(j);
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}
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}
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newWord += lastLetter;
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words[i] = newWord;
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}
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}
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System.out.print("\nFinal string: ");
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System.out.println(String.join(" ", words));
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} catch (Exception e) {
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System.out.println("An error occurred: " + e.getMessage());
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}
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}
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}
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