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
Related
I'm still learning Java and me and my classmate are working on this assignment code game hangman. We have a problem in the part printing the original word, because we want the word to be hidden, but we need the word to be printed, not hidden on the final output. What can we revise in our code to make it print properly??
import java.util.Scanner;
public class GameHangman {
public static void main(String[] args) {
System.out.println("Welcome! To the HANGMAN game.");
System.out.println("Guess the word by guessing each letter.");
System.out.println("LET'S START!");
String[] words = {"superman","batman","spiderman"};
int t = words.length;
int x, y, miss;
String w;
char l, c='y';
Scanner input = new Scanner(System.in);
while (c=='y')
{
x = (int)(Math.random()*(t));
w = words[x];
char[]chars = w.toCharArray();
var hidden = new char[w.length()];
for(int i = 0; i < hidden.length; i++)
{
hidden[i] = '*';
}
boolean z = false;
int count = 0;
miss = 0;
while(!z)
{
System.out.print("\n(guess) Enter a letter ("+new String(hidden)+"): ");
l = input.next().charAt(0);
y = 0;
for(int i = 0; i < hidden.length; i++)
{
if(chars[i]==l)
{
y++;
if(hidden[i]=='*')
{
hidden[i]=l;
count++;
y++;
}
}
}
if(y==1)System.out.println("\nOOPS!("+l+") already present, try other letters.");
else if (y==0)
{
miss++;
System.out.println("\nYIKES!("+l+") is not in the word, guess again.");
}
if(count==hidden.length) z = true;
}
System.out.println("\nGreat!! You guessed the word("+hidden+")!!!");
System.out.println("You were wrong "+miss+" tries.");
System.out.print("\nDo you want to continue with another game word? "
+ "\nEnter yes or no: ");
c = input.next().charAt(0);
}
}
}
in this part, we print our final word
System.out.println("\nGreat!! You guessed the word("+hidden+")!!!");
and this is what we are getting output. we need to print with not being hidden.
Here are two methods to convert your char array hidden into a String for printing out.
// First option: Create a String object
String str1 = new String(hidden);
System.out.println("\nGreat!! You guessed the word("+str1+")!!!");
// Second option: Using valueOf method
String str2 = String.valueOf(hidden);
System.out.println("\nGreat!! You guessed the word("+str2+")!!!");
Alternatively, you can also print your String variable w and avoid calling these methods.
I wanted to make a program in which only repeats words that has 3 of the same letters back to back. eg the mooonkey raaan through the mounnntains. the program should only repeat mooonkey, raaan
public class Triplets2 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("write a sentence");
String in = input.nextLine();
String [] sentence = in.split(" ");
for (int i = 0; i < sentence.length; i++) {
char [] word = sentence[i].toCharArray();
int counter =0;
for (int s = 0; s < word.length; s++) {
char letter = word[s];
for (int x = 0; x<word.length; x++) {
if (letter == word[x]) {
counter++;
}
else {
counter = 0;
}
}
}
if (counter >=3) {
System.out.print(sentence[i] + ", ");
}
}
}
the program instead just repeats nothing.
Your code is almost correct, the only logical error you made is inside your inner loop you keep resetting your counter variable as soon as you find a letter that is different:
if (letter == word[x]) {
counter++;
} else {
counter = 0;
}
So when you iterate over a word like "raaan" your counter will reset when it reaches the very end of the String, because "n" only exists once.
What this means is that you will only be able to detect words that have 3 consecutive letters at the very end (like "Hooo").
The solution is simple:
Once you found 3 consecutive letters in a word you can just stop iterating and checking the rest of your word. At that point you already know that it fits your criteria:
if (letter == word[x]) {
counter++;
if(counter >= 3) break; // stop inner loop checking once we found 3 letters
} else {
counter = 0;
}
Since you are looking for consecutive letters you want to start at char i and then compare the char at i to char at i+1 and at i+2. If they are all equal then we have a match and can continue.
You can simplify the whole function such as:
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("write a sentence");
String in = input.nextLine();
List<String> tripleLetter = new ArrayList<>();
for (String s : in.split(" ")) {
char[] word = s.toCharArray();
for (int i = 0; i < word.length - 2; i++) {
if ((word[i] == word[i+1]) && (word[i] == word[i+2])) {
tripleLetter.add(s);
break;
}
}
}
System.out.println(tripleLetter.stream().collect(Collectors.joining(", ")));
}
Allow me to suggest a solution that differs slightly from yours and doesn't use a counter.
Scanner input = new Scanner(System.in);
System.out.println("write a sentence");
String in = input.nextLine();
String[] sentence = in.split(" ");
for (int i = 0; i < sentence.length; i++) {
char[] word = sentence[i].toCharArray();
for (int s = 0; s < word.length - 2; s++) {
if (word[s] == word[s + 1] && word[s] == word[s + 2]) {
System.out.print(sentence[i] + ", ");
break;
}
}
}
Check whether the current letter, in the current word, is the same as the next letter and the same as the letter after the next letter. If the condition holds, then print the current word and proceed to the next word in the sentence.
Well, if you're just looking for a shorter version of doing this then try this.
first, split the sentence on one or more white space characters (you should be doing that regardless).
stream the array and filter on a single character, followed by the same two characters via a back reference to the capture group (see regular expressions for that).
And print them.
String str =
"Thiiis is aaaa tesssst of finding worrrrds with more than threeeeee letteeeeers";
Arrays.stream(str.split("\\s+"))
.filter(s -> s.matches(".*(.)\\1\\1.*"))
.forEach(System.out::println);
Prints
Thiiis
aaaa
tesssst
worrrrds
threeeeee
letteeeeers
I have created a palindorme java program which is getting an error.the error is saying int cannot be converted to boolean.
import java.util.Scanner;
public class palindrome
{
public static void main()
{
Scanner sc = new Scanner(System.in);
int l,i;
String s,s1;
System.out.println("Enter your string");
s = sc.nextLine();
l = s.length();
for(i=0;l-i-1;i++)
{
s1 = s + s.charAt(i);
}
if(s1==s)
System.out.println("This is Palindrome");
else
System.out.println("This is not a Palindrome");
}
}
For loop condition seems wrong.
for(initial counter; condition to terminate; increase counter) {}
for(i=0; i<l; i++) {}
Along with the answer above you can try a different approach. You don't need to go all the string length to check a palindrome. A palindrome can be checked iterating half of the array length like this -
public void checkPalindrome(String strToCheck){
char[] arr = strToCheck.toCharArray();
int size = arr.length;
char [] original = Arrays.copyOf(arr,arr.length);
for (int i = 0; i < size / 2; i++) {
char temp = arr[i];
arr[i] = arr[size-i-1];
arr[size-i-1] = temp;
}
if(Arrays.equals(arr, original)) {
System.out.println("Palindrome");
} else {
System.out.println("Not a palindrome");
}
}
What are done here:
reversing the string first iterating the halfway
comparing the reversed string with the original using Arrays.equals() method.
There are quite a few things off here, first here is the fixed code:
public static void main(String [] args)
{
Scanner sc = new Scanner(System.in);
int l,i;
String s = "",s1 = "";
System.out.println("Enter your string");
s = sc.nextLine();
l = s.length();
for(i = l - 1; i >= 0; i--)
{
s1 = s1 + s.charAt(i);
}
if(s1.equals(s))
System.out.println("This is Palindrome");
else
System.out.println("This is not a Palindrome");
}
The first thing to fix was your for loop, as you saw you were getting an error. This was fixed by setting the initial i to the length minus 1, changing the loop condition to i >= 0, and using i-- to subtract 1 from i each loop.
These changes to the loop were made so that the character starting from the last position in the String is the first one being return by s.charAt(i) so you can reverse the String. I think you were attempting to do something along these lines to add the characters starting from the end to a String.
I also changed s1 = s + s.charAt(i) to s1 = s1 + s.charAt() so the correct String is being appended. (This should probably be StringBuilder however).
s and s1 now have the initial condition of "" instead of nothing.
And finally you cannot compare String equality with ==, it must be s1.equals(s).
Test Run:
Enter your string
racecar
This is Palindrome
I'm scanning 2 strings during each iteration and storing it in s and t. Only during the first iteration, the first string that I scan is getting stored in t and not in s (I got to know this by debugging in eclipse). During successive iterations the piece of code works fine. I'm not able to understand what is going on during the first iteration. Please help me. Thanks.
import java.io.*;
import java.util.*;
public class ResidentInfo {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner scan = new Scanner(System.in);
int i,n;
n = scan.nextInt();
for(i=0 ; i<n ; i++)
{
int sl,tl,j,k;
String s, t;
boolean flag = false;
s = scan.nextLine();
t = scan.nextLine();
sl = s.length();
tl = t.length();
char[] sa = new char[sl];
char[] ta = new char[tl];
sa = s.toCharArray();
ta = t.toCharArray();
for(j=0 ; j<sl ; j++)
{
for(k=0 ; k<tl ; k++)
{
if(sa[j]==ta[k])
{
flag = true;
break;
}
}
if(flag)
{
break;
}
}
if(flag)
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
}
}
}
The first thing this code does is determine if the two strings' lengths are equal. If the strings' lengths are not equal, the code prints no. If the lengths are equal, the code then checks each character in the exact same index in the strings, to see if they are equal. If the characters at a specific index is not equal, the code breaks the loop and prints no. If all of the characters at each index are equal the code prints yes.
You asked a question about next() and nextLine().
Try: Question asked already.
Scanner scan = new Scanner(System.in);
System.out.print("enter a number: ");
int n = scan.nextInt();
for(int i=0; i < n; i++)
{
boolean flag = true;
System.out.print("enter something for s: ");
String s = scan.next();
System.out.print("enter something for t: ");
String t = scan.next();
if(s.length() == t.length())
{
for(int j = 0; j < s.length(); j++)
{
if(s.charAt(j) != t.charAt(j))
{
flag = false;
break;
}
}
if(flag)
{
System.out.println("Yes");
}
else
{
System.out.println("NO");
}
}
else
{
System.out.println("NO");
}
I have been working on my hangman program for far to long and cannot figure out why it is not replacing the characters entered with the asterisks.
There are a lot of details I have not added so please do not sit here and judge that. I need someone to tell my why the character the user enters is not replacing the asterisks and If you know what I could do to fix it please tell me.
I'm struggling. I have edited my program to show you where I know the logic error is coming from however I do not know what the error is.
String hiddenWord = wordList[rand];
char[] asterisks = new char[MAXCHAR];
hideWord(hiddenWord);
System.out.println(Arrays.toString(hideWord(hiddenWord)));
numGuess( hiddenWord,asterisks);
public static char[] hideWord(String hiddenWord)
{
int wordLength = hiddenWord.length();
//int length = wordLength * 2;
char[] asterisks = new char[wordLength];
for(int i=0; i < wordLength; i++)
{
asterisks[i] = '*';
}
return asterisks;
}
public static void numGuess(String hiddenWord,char[] asterisks)
{
Scanner keyboard = new Scanner(System.in);
hideWord(hiddenWord);
int remAttempts = MAXGUESS;
int i = 0;
while(i < (hiddenWord.length()-1))
{
System.out.println("Enter a letter or 9 to quit");
char guess = keyboard.next().charAt(i);
if(asterisks[i] == (hiddenWord.charAt(i)))
{
//attemtps == hiddenWord.charAt(i);
System.out.println("Nice job!");
remAttempts--;
}
i++;
}
}
Look at this code (I changed the formatting a bit):
while (i < hiddenWord.length() - 1) {
System.out.println("Enter a letter or 9 to quit");
char guess = keyboard.next().charAt(i);
//...
i++;
}
You're asking for a letter, but you really request a String with at least the size + 1 that equals i: keyboard.next().charAt(i);. Therefore, if you write just a letter, then you'll get an Exception at the second iteration of that loop.
I guess what you meant was: keyboard.next().charAt(0);. This will return the first character of the given String.
If this doesn't solve the problem, then provide the whole Stacktrace and mark the line in your code, where the Exception occurs.