While creating a word counter in java through collecting user input from a scanner I've been running into the error of having my program display that there was one extra word entered when there is a space entered in after the final word in the char. Is there a way to check the user input for a space and remove it before entering my word count loop?
Right after...
String userinput = wordcounter.nextLine();
... add this line:
userinput = userinput.trim();
It should fix the problem, since the function trim() gets rid of any blank spaces at the beginning or ending of the string.
Try something like this:
public static void main(String[] args) {
int word = 0;
System.out.println("Enter a string: ");
Scanner wordcounter = new Scanner(System.in);
while (wordcounter.hasNext()) {
wordcounter.next();
word++;
}
System.out.println("You hav entered " + word + " words.");
}
Moreover there are many functions in Java that you can use here to do this very easily like using string tokenizer, etc. But if you want to make changes in the above program, you could simply do a simple check statement before the main if condition that if the I == length-1, because this means that you are at the last character and there is nothing after this, for this case you could just continue or break. Hopefully I solved your query, do tell in case you need more clarification...
One possible way: use regex after you trim your input.
String text = "I turned myself into a pickle Morty! :)";
String[] words = text.split("([\\W\\s]+)");
int count = 0;
for (String word: words) {
count++;
}
System.out.println("You have entered " + count + " words!");
Scanner in = new Scanner (System.in);
System.out.println("Write something");
String x = in.nextLine();
char y;
char z;
int n =1;
boolean t = true;
boolean f = false;
boolean o = false;
for (int i = 0; i<x.length();i++){
y = x.charAt(i);
if(t){
if(' ' == y){n--;}
else if(y != ' '){ t=false; f = true;}
}
if(' ' == y){n++;}
if(o){
z = x.charAt(i-1);
if(' ' == y && ' '== z){n--;}
}
if (o){
if(' ' == y && i == x.length()-1){n--;}
}
if(f){o = true;}
}
System.out.println(n);
This actually works lol
Related
I am a student and kind of new to Java. For my homework I have to:
Ask the user to input a number (at least 7) using a do while loop.
Using a for loop I am required to ask the user to input that number of words.
Then I have to check if one of the words fulfills the given conditions:
The word must:
Start with an uppercase letter
End with a number
Contain the word "cse".
I am asked to create a method inside some code homework that does a specific task, the method should check all the required conditions, the name of the method should be countTest and it accepts the String as a parameter.
I will show you my code but I don't know how to create this specific method.
Output format
System.out.println("There as a total number of words " + count + " and
the ones that fulfill the condition are: " + condition);
The problem is, I dont know how to create the method or constructor or whatever it is called that calls all of the 3 methods inside it, and then connect that particular method to the main method!
I hope you guys can understand I am new to this, thank you in advance!
public class D6_6 {
public static void main(String[]args){
Scanner sc = new Scanner(System.in);
System.out.println("Type a number that is at least 7");
int number = sc.nextInt();
int count = 0;
int condition = 0;
do{
if(number<7){
System.out.println("You should type a number that is at least 7 or higher");
number = sc.nextInt();
}
}
while(number<7);
sc.nextLine();
String str;
for(int i =0; i<number; i++){
System.out.println("Type a word");
str = sc.nextLine();
count++;
}
}
public boolean countTest(String str) {
}```
To check if the word start with an uppercase:
You can do that by first selecting the character you want to check by str.charAt(0). This will return a char that is the first letter of the input str.
To check if this char is an uppercase letter, you can easily use char.isUppercase(). This will return a boolean. You have to replace char by the name of the variable were you put the char of str.charAt(0) in.
To check if the last character is a number:
You can do that again by first selecting the last character by str.charAt(str.length()-1), were string.length-1 is the number of the last character.
To check if this character is a number, you can use the ascii table. Every character has it's own number. So if you want to check if your character is between 0 and 9, you can use char >= 48 || char <= 57 (look up in the ascii table). Again, char is the name of the variable were you put the char of str.charAt(str.length()-1) in.
To check if the word contains "cse":
There is a very easy method for that: str.contains("cse") will return a boolean that is true when "cse" is in the word and false when the word does not contain "cse".
I hope it is clear for you now!
I think I did it, thank you guys very much, I appreciate it!
public class D6_6 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Type a number that is at least 7");
int number = sc.nextInt();
int count = 0;
int condition = 0;
do {
if (number < 7) {
System.out.println("You should type a number that is at least 7 or higher");
number = sc.nextInt();
}
}
while (number < 7);
sc.nextLine();
String str;
for (int i = 0; i < number; i++) {
System.out.println("Type a word");
str = sc.nextLine();
count++;
if((countTest(str))){
condition++;
}
}
if(count == 0){
System.out.println("No words typed");
} else {
System.out.println("Total number of words typed: " + count + ", which fulfill the condition: "+ condition);
}
}
public static boolean countTest(String str) {
return Character.isUpperCase(str.charAt(0)) && str.charAt(str.length() - 1) >= 48 || str.charAt(str.length() - 1) <= 57 || str.contains("cse");
}
}```
The following Java program is supposed to manipulate a string input by the user in such a way that the user will decide which character needs to be replaced with another and just the last character from the string should be replaced. Example if the user enters the string "OYOVESTER" and decides to replace "O" with "L", the program should output the following result: "OYLVESTER" (notice that only the last "O" was replaced with "L")
NOTE: YOU CANNOT USE BREAK COMMAND TO STOP THE LOOP. IT IS PROHIBITED.
import java.util.Scanner;
public class StringFun {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter the string to be manipulated");
String inString = keyboard.nextLine();
String outString = "";
//Replace Last
System.out.println("Enter the character to replace");
char oldCharF = keyboard.next().charAt(0);
System.out.println("Enter the new character");
char newCharF = keyboard.next().charAt(0);
int count = 0; // variable that tracks number of letter occurrences
for(int index = inString.length() - 1;index >= 0;index--) {
if(inString.charAt(index) == oldCharF && count < 1){
outString = newCharF + outString;
outString = outString + inString.substring(0,index);
count++;
}
if (count < 1) {
outString = outString + inString.charAt(index);
}
}
System.out.print("The new sentence is: "+outString);
}
}
I keep getting the following output which is incorrect:
Enter the string to be manipulated
OYOVESTER
Enter the character to replace
O
Enter the new character
L
The new sentence is: LRETSEVOY
There are many simpler ways to achieve your requirement but I hope you have to demonstrate this with loops (without breaks)
Then you can use some thing like this :
boolean skip = false;
for (int index = inString.length() - 1; index >= 0; index--) {
if (!skip && inString.charAt(index) == oldCharF) {
outString = newCharF + outString;
skip = true;
}
else {
outString = inString.charAt(index) + outString;
}
}
PS : Using String concatenation inside loops is not recommended since
every String concatenation copies the whole String, usually it is preferable to
replace it with explicit calls to StringBuilder.append() or StringBuffer.append()
No break command seems like a weird condition. You could just a boolean value, and other methods, to break the loop when you need. Why not do something like this?
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter the string to be manipulated");
String word = keyboard.nextLine();
//Replace Last
System.out.println("Enter the character to replace");
char oldCharF = keyboard.next().charAt(0);
System.out.println("Enter the new character");
char newCharF = keyboard.next().charAt(0);
int index = word.lastIndexOf(oldCharF);
if(index > 1){
word = word.substring(0,index) + newCharF + word.substring(index+1);
}
System.out.println("The new sentence is: " + word);
}
I am new to java and am trying to create a palindrome word program, to check if the word backwards is the same.
public static void isPalindromeWord(){
Scanner input = new Scanner (System.in);
System.out.print("Enter a word to check: ");
String word = input.next().toLowerCase();
String reversed = new StringBuffer(word).reverse().toString();
int len = word.length();
for(int x = 0; x < len ; x++){
if(word.charAt(x) == reversed.charAt(x)){
System.out.println("True");
}else{
System.out.println("False");
}
}
}
Please excuse if I've done anything wrong, I have only started learning today.
My problem is :
With the current it outputs True for something such as "otto" which is a palindrome. But it also does True for "oplko" which isn't. So I know that it only checks the first and last letters but I thought that with the for loop it will go through each letter?
Can someone be kind enough to explain where I am going wrong and suggest how to fix it? The reason I am using a for loop is because the task is requiring me to do so.
You are very close to the solution. Since you have already reversed the string you can check if they are equal
new StringBuffer(word).reverse().equals(word);
Edit: Added one more solution for using loop
What you are doing in the loop is mostly correct. You are getting True for oplko is because you are not exiting the loop when the word.charAt(x) == reversed.charAt(x) condition fails. This can be fixed by
public static void isPalindromeWord() {
Scanner input = new Scanner(System.in);
System.out.print("Enter a word to check: ");
String word = input.next().toLowerCase();
String reversed = new StringBuffer(word).reverse().toString();
int len = word.length();
for (int x = 0; x < len; x++) {
if (word.charAt(x) != reversed.charAt(x)) {
System.out.println("False");
return;
}
}
System.out.println("True");
}
there are a lot of ways to do what you want (including Anthony C's very elegant answer), but here is a simple fix to make yours work :
public static void isPalindromeWord(){
Scanner input = new Scanner (System.in);
System.out.print("Enter a word to check: ");
String word = input.next().toLowerCase();
//you don't really need to get a reverse here
//String reversed = new StringBuffer(word).reverse().toString();
int len = (int)(word.length() / 2);//only check half (and not evnt the middle one for odd numbers
boolean isPalindrom = true;
for(int x = 0; x < len ; x++){
if(word.charAt(x) != word.charAt(word.length() - 1 - x)){
isPalindrom = false;
//at least one difference, this is not a palindrome
break;
}
}
if(isPalindrom)
System.out.println("True");//the for wasn't broken
else
System.out.println("False");
}
public static void isPalindromeWord(){
Scanner input = new Scanner (System.in);
System.out.print("Enter a word to check: ");
String word = input.next();
String reversed = input.next();
char c[]=word.toLowerCase().toCharArray();
char d[]=reversed.toLowerCase().toCharArray();
if(word.length() != reversed.length()) System.out.print("False ");
Arrays.sort(c);
Arrays.sort(d);
if(Arrays.equals(c,d)) System.out.print("True ");
else System.out.print("False ");
}
Just to describe where you have a mistake. You are writing in console on every step of loop, doesn't matter which result of comparing do you have.
boolean isPolindrom = true;
for(int x = 0; x < len ; x++){
isPolindrom = (word.charAt(x) == reversed.charAt(x)); //compare chars and store result into variable
if (!isPolindrom) { //if chars are different break a loop, because word is already not a polindrom
break;
}
}
System.out.println(isPolindrom); //output result
Solution from #AnthonyC is really better
import java.util.*;
public class VowelCounter
{
public static void main(String[] args)
{
Scanner keyboard = new Scanner(System.in);
System.out.println("Input a series of characters: ");
String letters = keyboard.next();
int count = 0;
for (int i = 0; i < letters.length(); i++)
{
char characters = letters.charAt(i);
if (isVowel(characters) == true)
{
count++;
}
}
System.out.println("The number of vowels is: " + count);
}
public static boolean isVowel(char characters)
{
boolean result;
if(characters=='a' || characters=='e' || characters=='i' || characters=='o' || characters=='u')
result = true;
else
result = false;
return result;
}
}
The code works but im suppose to input "Spring break only comes once a year." which if i do with the spaces my program will only find the vowels of Spring. how do i make it so it will skip the spaces and read the whole sentence.
This is your problem:
String letters = keyboard.next();
It has nothing to do with the vowel-counting part - but everything to do with reading the value. The Scanner.next() method will only read to the end of the token - which means it stops on whitespace, by default.
Change that to
String letters = keyboard.nextLine();
and you should be fine.
You should verify this is the problem by printing out the string you're working with, e.g.
System.out.println("Counting vowels in: " + letters);
When you do:
String letters = keyboard.next();
The Scanner stops reading at the first whitespace.
To read the complete phrase until you press enter, you should use nextLine() instead:
String letters = keyboard.nextLine();
Just use
String letters = keyboard.nextLine();
instead of
String letters = keyboard.next();
This is because .nextLine() will read line by line so that you can have your complete statement in latters. Hope this will help you
I'm trying to write my own Java word count program. I know there may already be a method for this, but I'd like to get it work. I'm getting an out of bounds error at line 14. I'm trying to use an input word to count how many times it appears in an input string. So I'm looping up to stringlength - wordlength, but that's where the problem is.
Here is the code:
import java.util.Scanner;
public class wordcount {
public static void main(String[] args)
{
Scanner s = new Scanner(System.in);
System.out.print( "Enter word : " );
String word = s.nextLine();
Scanner t = new Scanner(System.in);
System.out.print("Enter string: ");
String string = t.nextLine();
int count = 0;
for (int i = 0; i < string.length()-word.length(); i = i+1){
String substring = string.substring(i,i+word.length());
if (match(substring, word)==true){
count += 1;
}
}
System.out.println("There are "+count+ " repetitions of the word "+word);
}
public static boolean match(String string1, String string2){
for (int i=0; i<string1.length(); i+=1){
if (string1.charAt(i)!=string2.charAt(i)){
return false;
}
}
return true;
}
}
First of all, two Scanners are not necessary, you can do many inputs with the same Scanner object.
Also, this if condition
if (match(substring, word) == true)
can be rewritten like
if (math(substring, word))
I would also recommend you to use i++ to increase the loop variable. Is not strictly necessary but is "almost" a convention. You can read more about that here.
Now, about theIndexOutOfBoundsException, I've tested the code and I don't find any input samples to get it.
Besides, there is an issue, you are missing one iteration in the for:
for (int i = 0; i < string.length() - word.length() + 1; i++) { // Add '+ 1'
String substring = string.substring(i, i + word.length());
// System.out.println(substring);
if (match(substring, word)) {
count++;
}
}
You can test it by putting a print statement inside the loop, to print each substring.
I'm not getting an out of bounds error, can you tell me what values you were using for word and string?
I have identified a bug with your program. If word is equal to string, it still returns count 0. I suggest adding one more iteration and using regionMatches instead. RegionMatches makes your match method obsolete and will return false if word.length() + i is equal or greater than string.length(), avoiding out of bounds issues.
As you can see I also moved the calculations to a seperate method, this will make your code more readable and testable.
And as Christian pointed out; you indeed do only need one Scanner object. I've adapted the code below to reflect it.
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter word : ");
String word = sc.nextLine();
System.out.print("Enter string: ");
String string = sc.nextLine();
int count = calculateWordCount(word, string);
System.out.println("There are " + count + " repetitions of the word " + word);
}
private static int calculateWordCount(String word, String string) {
int count = 0;
for (int i = 0; i < string.length() - word.length() + 1; i++) {
if (word.regionMatches(0, string, i, word.length())) {
count++;
}
}
return count;
}