The problem is in the program's action, if the last letters in the input data in the line

In the loop, you form a string of letters that you want to draw. The conversion itself occurs only under the condition that the next character is a space or a digit. So if the line ends with letters and it comes to an end - there is no appeal (and indeed, the output of this part). To fix this, you can add a processing after the cycle with the condition that there are not displayed characters.

 if (!first.isEmpty()) { String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; } 

I will try to write a detailed answer indicating the shortcomings and possible solutions to them.

For starters, your code is only formatted to a more readable form. For more information, read the Google Java Style Guide .

 import java.util.Scanner; public class Anagramm { public static void main(String args[]) { Scanner scanner = new Scanner(System.in); // это сканнер System.out.println("Введите строку"); String string = scanner.nextLine(); String first = new String(); // стек строки первичных данных int stringLenght = string.length(); // значение длины строки для счетчика цикла for (int i = 0; i < stringLenght; i++) { char myChar = string.charAt(i); char type = simbolType(myChar); // тип анализируемого символа, отсыл к методу simbolType if (type == 'l') { first = first + myChar; // образование первичной строки, прямой } else if (type == 'd') { String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; int z = Character.getNumericValue(myChar); System.out.print(z); // print digit } else if (type == 'w') { String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; System.out.print(" "); } else { String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; System.out.print(myChar); } // условие для формирования блока для анаграммы char k = type; if (k != 'l') { String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; } else {} } } public static char simbolType(char unknownSymbol) { // блок определения типа символа char hisType; if (Character.isLetter(unknownSymbol) == true) { hisType = 'l'; // тип символа буква } else if (Character.isDigit(unknownSymbol) == true) { hisType = 'd'; // тип символа цифра } else if (Character.isWhitespace(unknownSymbol) == true) { hisType = 'w'; // тип символа пробел } else { hisType = 'a'; // тип символа другой "any" } return hisType; } } 

The variable stringLenght used once in the loop condition, omit its definition and calculate the length of the string directly in the loop condition.

 for (int i = 0; i < string.length(); i++) { ... } 

Code formatting the output of the result, consisting of blocks of conditions if , else if , else , can be rewritten using the switch , so that the code will look cleaner.

 switch (type) { case 'l': first = first + myChar; // образование первичной строки, прямой break; case 'd': String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; int z = Character.getNumericValue(myChar); System.out.print(z); // print digit break; case 'w': String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; System.out.print(" "); break; default: String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; System.out.print(myChar); } 

Some actions in the d , w , default blocks are repeated, you can optimize and combine them.

 switch (type) { case 'l': first += myChar; // образование первичной строки, прямой break; case 'd': case 'w': default: String r = new StringBuffer(first).reverse().toString(); first = ""; System.out.print(r); System.out.print(myChar); } 

In this piece of code, nothing happens, because at the entrance to the if variable first always empty. You can safely delete it.

 char k = type; if (k != 'l') { String r = new StringBuffer(first).reverse().toString(); System.out.print(r); first = ""; } else {} 

As a result, we received such a code.

 import java.util.Scanner; public class Anagramm { public static void main(String args[]) { Scanner scanner = new Scanner(System.in); // это сканнер System.out.println("Введите строку"); String string = scanner.nextLine(); String first = new String(); // стек строки первичных данных for (int i = 0; i < string.length(); i++) { char myChar = string.charAt(i); char type = simbolType(myChar); // тип анализируемого символа, отсыл к методу simbolType switch (type) { case 'l': first += myChar; // образование первичной строки, прямой break; case 'd': case 'w': default: String r = new StringBuffer(first).reverse().toString(); first = ""; System.out.print(r); System.out.print(myChar); } } } public static char simbolType(char unknownSymbol) { // блок определения типа символа char hisType; if (Character.isLetter(unknownSymbol) == true) { hisType = 'l'; // тип символа буква } else if (Character.isDigit(unknownSymbol) == true) { hisType = 'd'; // тип символа цифра } else if (Character.isWhitespace(unknownSymbol) == true) { hisType = 'w'; // тип символа пробел } else { hisType = 'a'; // тип символа другой "any" } return hisType; } }