I'm trying to write a program which will output what Palindromes will work from entering in a string and how many there are. I keep getting a lot of errors and I'm still trying to get my head around some of the harder topics in Java!
Here's what I have already, as always, all answers are greatly appreciated!
public static boolean Palindrome(String text) {
int index;
int palindrome;
System.out.println("Please enter your text ");
text = EasyIn.getString();
for(index = 0; index < amount.length() / 2; index++) {
if(text.charAt(index) != text.charAt(text.length() - index - 1)) {
return false;
}
}
System.out.println("The number of valid palindrome(s) is " + amount);
amount = EasyIn.getString();
}
I think the problem is in amount.length(), you should use text.length(), since you are looping over the half of text. The algorithm works fine. Here is a reduced example:
public static boolean palindrome(String text)
{
for (int index = 0; index < text.length() / 2; index++) {
if (text.charAt(index) != text.charAt(text.length() - index - 1)) {
return false;
}
}
return true;
}
Note:
You forgot to add a return true statement, if you don't add one, is possible that the for loop finishes and no return statement is reached, which will cause an error.
I would recommend you to follow Java naming conventions. You method should be called like someMethodName instead of SomeMethodName. This last is used for class names.
Edit:
As #bobbel commented, you could improve this code by assigning text.length() to a variable and using it inside the for.
There can be two things:
ammount variable that you used either it could be a string array that you are maintaining strings inside it, if this is the case than you have to loop first through array of strings and then maintain one nested loop to check that strings inside it are palindrom or not
or second case is that you have used the variable incorrect it may be text instead of ammount
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("enter string to check for palidrome");
String orginal = in.next();
int start = 0;
int middle = orginal.length()/2;
int end = orginal.length() - 1;
int i;
for(i=start; i<=middle; i++) {
if(orginal.charAt(start) == orginal.charAt(end)) {
start++;
end--;
} else {
break;
}
}
if(i == middle+1) {
System.out.println("palidrome");
} else {
System.out.println("not palidrome");
}
}
This is the Simplest way of Checking Palindrom number.
package testapi;
public class PalindromNumber {
public static void checkPalindrom(Object number) {
StringBuilder strNumber = new StringBuilder(number.toString());
String reverseNumber = strNumber.reverse().toString();
if (number.toString().equals(reverseNumber)) {
System.out.println(number + " is palindrom number");
} else {
System.out.println(number + " is not palindrom number");
}
}
public static void main(String[] args) {
checkPalindrom(101);
checkPalindrom(10.01);
checkPalindrom("aanaa");
}
}
Related
I've been trying to create an algorithm where each letter adds points. I don't want to use charAt, I'd like to use the substring method.
My problem is that String letter does not seem to get each letter and the result is always 0.
Is there a way to get each letter and convert it to points?
public class WDLPoints{
public static void main(String[] args){
String word = "LDWWL";
System.out.println(getMatchPoints(word));
}
public static int getMatchPoints(String word) {
int points = 0;
String letter = word.substring(5);
for (int i = 0; i < word.length(); i++) {
if (letter.equals("W")) {
points+=3;
}
else if (letter.equals("D")) {
points+=1;
}
else {
points = 0;
}
}
return points;
}
}
You may try the following changes in your public static int getMatchPoints(String word) method:
for (int i = 0; i < word.length(); i++) {
String letter = word.substring(i, i + 1);
if (letter.equals("W")) {
points+=3;
}
else if (letter.equals("D")) {
points+=1;
}
}
word.substring(i, i + 1) will get a single letter word and will help you compute your score the way you want.
If you want to make it really simple you can just use String.toCharArray() and then iterate over the array of char and check its value:
public static int getMatchPoints(String word) {
int points = 0;
char[] arr = word.toCharArray();
for (char letter : arr) {
if (letter == 'W') {
points += 3;
}
else if (letter == 'D') {
points += 1;
}
}
return points;
}
I also removed your else statement because that was just setting the value to 0 if there is any other letter in the loop. I think you intended it to be points += 0 which does nothing, so it can just be removed.
Example Run:
Input:
String word = "LDWWL";
Output:
7
Note: I am aware you might not be allowed to use this solution, but I thought it would be good info on the possibilities since it does not technically use charAt()
Also I'd like to point out you misunderstand what substring(5) does. This will return all characters after the position of 5 as a single String, it does not separate the String into different characters or anything.
You will find that your variable letter is always the empty String. Here's a better way of doing things:
class WDLPoints
{
public static void main(String[] args)
{
String word = "LDWWL";
System.out.println(getMatchPoints(word));
}
// We have only one method to encode character values, all in one place
public static int getValueForChar(int c)
{
switch((char)c)
{
case 'W': return 3;
case 'D': return 1;
default: return 0; //all non-'W's and non-'D's are worth nothing
}
}
public static int getMatchPoints(String word)
{
// for all the characters in the word
return word.chars()
// get their integer values
.map(WDLPoints::getValueForChar)
// and sum all the values
.sum();
}
}
Assuming your string represents a football teams performance of the last 5 games, you could keep it simple and readable with something like:
public static int getMatchPoints(String word) {
String converted = word.replace('W', '3').replace('D', '1').replace('L', '0');
return converted.chars().map(Character::getNumericValue).sum();
}
This converts your example input "LDWWL" to "01330" and sums each char by getting its numeric value.
Currently trying to program a poem Palindrome checker. This is not for palindromes specifically, but that the array has words in the same order both ways. For example the following is a poem palindrome
Life-
imitates nature,
always moving, traveling continuously.
Continuously traveling, moving always,
nature imitates
life
My issue is iterating through the array to match the first and last elements, as currently it compares things in the wrong order.
My code is as follows:
import java.util.Scanner;
import java.io.*;
public class WordPalindromeTest {
public static void main(String[] args) {
System.out.println("This program determines if an entered sentence/word poem is a palindrome.");
Scanner input = new Scanner(System.in);
System.out.println("Please enter a string to determine if it is a palindrome: ");
while(input.hasNextLine()) {
String palin = input.nextLine();
if(palin.equals("quit")) {
break;
}
else {
boolean isPalin = isWordPalindrome(palin);
if(isPalin == true) {
System.out.println(palin + " is a palindrome!");
}
else
System.out.println(palin + " is NOT a palindrome!");
}
}
System.out.println("Goodbye!");
input.close();
}
public static boolean isWordPalindrome(String s) {
boolean isWordPal = false;
String lowerCase = s.toLowerCase();
String replaced = lowerCase.replaceAll("[^a-zA-Z0-9\\s]", "");
String words[] = replaced.split(" ");
for(int i = 0; i < words.length; i++) {
for(int j = 0; j < words.length; j++) {
if (words[i].equals(words[j]) && i != j) {
isWordPal = true;
}
else
isWordPal = false;
}
}
return isWordPal;
}
}
With the specific point in question being
public static boolean isWordPalindrome(String s) {
boolean isWordPal = false;
String lowerCase = s.toLowerCase();
String replaced = lowerCase.replaceAll("[^a-zA-Z0-9\\s]", "");
String words[] = replaced.split(" ");
for(int i = 0; i < words.length; i++) {
for(int j = 0; j < words.length; j++) {
if (words[i].equals(words[j]) && i != j) {
isWordPal = true;
}
else
isWordPal = false;
}
}
return isWordPal;
}
I am confused on how to properly set up the loop to compare the right elements. It should compare the first element to the last, the second to the second to last, etc. until the loop is finished. I realize I have it compare the first to the entire array before moving on.
This seems like a homework assignment so I won't give you a working solution. But this of it like this:
-You don't need two loops. You only need to compare the first to the last, the second to the second to last, etc. (Hint: if you subtract i-1 from the length of the Array you'll get the corresponding element to i that you need to compare to). Also you only need to iterate over half of the length of the Array
-If ever isWordPal becomes false, you need to return false. Otherwise it might get overwritten and at the end it will return true.
A string is a palindrome if it is spelled the same way backward and
forward.
Examples of palindromes include “Radar” and “Dammit, I’m mad!”.
Write a java program, PalindromeTester, that asks the user to enter a
word or sentence and then checks whether the entered string is a
palindrome or not.
Spaces, nonalphabetics (.,!:?-()\";), and case within the string have
to be ignored e.g., "Drab as a fool, aloof as a bard." is a
palindrome.
Your implementation should define and use the method isPalindrome to
test if a certain string is a palindrome. The signature of the
isPalindrome method is as follows:
boolean isPalindrome(String)
Following is a sample run of the program. The user’s input is shown in bold.
java PalindromeTester
Introduction to Computer Programming (CMPS 200)
Spring 2015-16 2 of 3
Enter a string: I love CMPS 200
The string "I love CMPS 200" is NOT a palindrome.
This is the code I made, it keeps giving me an error.
I would like to know what my error is and whether there's a faster easier way of writing this code
import java.util.Scanner;
public class PalindromeTester {
public static void main (String args []) {
Scanner console = new Scanner(System.in);
System.out.println("Enter a string: ");
String palindrome = console.next();
if (isPalindrome (palindrome)) {
System.out.print("The string \""+palindrome+" is a palindrome.");
} else {
System.out.print("The string \""+palindrome+" is NOT a palindrome.");
}
}
public static boolean isPalindrome (String palindrome) {
int constant = 1;
for (int i = 0 ; i <= (palindrome.length()-1) ; i++) {
for (int z= (palindrome.length()-1);i >= 0; i--) {
if (palindrome.charAt(i) <'#'||'Z'<palindrome.charAt(i)&&palindrome.charAt(i)<'`'||'['<palindrome.charAt(i)&&palindrome.charAt(i)<'{') {
i=i+1;
}
if (palindrome.charAt(z)<'#'||'Z'<palindrome.charAt(z)&&palindrome.charAt(z)<'`'||'['<palindrome.charAt(z)&&palindrome.charAt(z)<'{') {
z=z+1;
}
if (palindrome.charAt(i)==(palindrome.charAt(z))) {
constant = constant * 1;
} else {
constant = constant * 0;
}
}
}
if (constant == 0 ) {
return false;
} else {
return true;
}
}
}
One approach would be to strip the non alpha characters out of the string. Then check if the string is the same as itself reversed (while upper case):
public static boolean isPalindrome(String palindrome) {
StringBuilder sanitisedString = new StringBuilder();
for(char c : palindrome.toCharArray()) {
if(Character.isLetter(c)) {
sanitisedString.append(c);
}
}
return sanitisedString.toString().toUpperCase().equals(sanitisedString.reverse().toString().toUpperCase());
}
Index out of range is caused by palindrome.charAt(z)) after z = z + 1;
Keep simple :
public static boolean isPalindrome(String palindrome)
{
palindrome = palindrome.replaceAll("\\W", ""); // remove all non word character
palindrome = palindrome.toLowerCase();
int size = palindrome.length();
int halfSize = size / 2;
for (int i = 0; i < halfSize; i++)
{
if(palindrome.charAt(i) != palindrome.charAt(size - i - 1))
return false;
}
return true;
}
Why not just make a new String and save the reversed source(String) in it.
public static boolean readstring(String s)
{
String b = "";
for (int i= s.length() -1; i >=0 ;i--)
{
b = b + s.charAt(i);
}
System.out.print(b +" and "+ s +" ");
return b == s || b.Equals(s);
}
EDIT: Hopefully this meets the requirements, by the way dont use the word "ignore" but "allow"
public boolean isPalindrome(String word) {
int backward = word.length() - 1;
for (int x = 0; x < word.length(); x++) {
if (word.charAt(x)!= word.charAt(backward--)) {
return false;
}
}
return true;
}
So I am trying to write a Method that will return to me the number of occurrences of a String in Another String. In this case it's finding the number of spaces in a String. It's as if indexOf() is not recognizing the spaces.
Here is my Method:
public int getNumberCardsDealt()
{
int count = 0;
int len2 = dealtCards.length();
int foundIndex = " ".indexOf(dealtCards);
if (foundIndex != -1)
{
count++;
foundIndex = " ".indexOf(dealtCards, foundIndex + len2);
}
return count;
}
Here is my application:
public class TestDeck
{
public static void main(String [] args)
{
Deck deck1 = new Deck();
int cards = 52;
for(int i = 0; i <= cards; i++)
{
Card card1 = deck1.deal();
Card card2 = deck1.deal();
}
System.out.println(deck1.cardsDealtList());
System.out.println(deck1.getNumberCardsDealt());
}
}
Note that I already have a Card Class and the deal method works.
Check the documentation of the indexOf method. You are using it wrong.
You should change the invocation
" ".indexOf(dealtCards);
To
dealtCards.indexOf(" ");
That is, invoking the method on the concerned string and passing to it the character you are looking for, not the other way around.
Moreover, your method would not calculate it correctly anyway, you should change it to something like:
public int getNumberCardsDealt() {
int count = 0;
int foundIndex = -1; // prevent missing the first space if the string starts by a space, as fixed below (in comments) by Andy Turner
while ((foundIndex = dealtCards.indexOf(" ", foundIndex + 1)) != -1) {
count++;
}
return count;
}
#A.DiMatteo's answer gives you the reason why your indexOf doesn't work currently.
Internally, String.indexOf is basically just iterating through the characters. If you're always just looking for a single character, you can trivially do this iteration yourself to do the counting:
int count = 0;
for (int i = 0; i < dealtCards.length(); ++i) {
if (dealtCards.charAt(i) == ' ') {
++count;
}
}
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 8 years ago.
I made a program to search for a certain string in another string and print Word found if the condition is true or print word not found if condition is false
The logic is as follows
enter word
length of word
for searching for letter [1]
if true
then for till length of word match with string to be searched
else continue loop
But I always get word not found no matter what the input, please help me over here!!!
The code is as follows :-
import java.util.Scanner;
class Search_i_String
{
public static void main(String args[])
{
int flag=0;
Scanner Prakhar=new Scanner(System.in);
System.out.println("Enter a String");
String ori=Prakhar.nextLine();
System.out.println("Enter the String to be Searched");
String x=Prakhar.nextLine();
char a[]=new char[ori.length()];
char b[]=new char[x.length()];
for(int i=0;i<ori.length();i++)
{
a[i]=ori.charAt(i);
}
for(int i=0;i<x.length();i++)
{
b[i]=x.charAt(i);
}
for(int i=0;i<a.length;i++)
{
if (a[i]==b[0])
{
for(int j=0;j<b.length;j++)
{
while(flag==0)
{
if(b[j]==a[i])
{
flag=0;
}
else if(b[j] != a[i])
{
flag=1;
}
i++;
}
}
}
}
if (flag==0)
{
System.out.println("Word Found !!!");
}
else
System.out.println("Word not Found");
}
}
P.S. : I know I can use the contains() function but I can as my professor suggests against it and could someone please correct the program I have written, because I could have scavenged off a program from the internet too if I had to, I just wanted to use my own logic
Thank You(again)
while(flag==0)
{
if(b[j]==a[i])
{
flag=0;
}
else if(b[j] != a[i])
{
flag=1;
}
i++;
j++; //add this and try once
}
If you are comparing strings in Java, you have to use equals();
So, stringA.equals(stringB);
Cheers!
Let me get this straight. You're looking for b array inside of a array? "Enter the string to be searched" means that you are searching the other way around, but I'll go with the logic your code seems to follow... Here's a naive way to do it:
if (a[i]==b[0])
{
flag = 0;
for(int j=0;j<b.length;j++)
{
if(b[j] != a[i+j]) // will array index out of bounds when its not foud
{
flag++; // you should probably break out of a named loop here
}
}
if(flag == 0){/*win*/}
}
You're modifying your first search loop with variable i when you don't have to. You can just add i to j. Also, you don't need the while loop inside if i'm understanding your problem. Like others have said, functions exist to do this already. This algorithm isn't even as efficient as it could be.
I know of an algorithm where you check starting in the last character in b instead of the first character in b to begin with. Then you can use that information to move your search along faster. Without resorting to full pseudo code, anyone know what that's called?
The simple way(but not the fastest way) is use double loop to check the chars in strings one by one, pls ref to my code and comments:
public class SearchString {
public static void main(String[] args) {
String a = "1234567890";
String b = "456";
// Use toCharArray() instead of loop to get chars.
search(a.toCharArray(), b.toCharArray());
}
public static void search(char[] a, char[] b) {
if (a == null || b == null || a.length == 0 || b.length == 0) {
System.out.println("Error: Empty Input!");
return;
}
int lenA = a.length, lenB = b.length;
if (lenA < lenB) {
System.out
.println("Error: search key word is larger than source string!");
return;
}
// Begin to use double loop to search key word in source string
for (int i = 0; i < lenA; i++) {
if (lenA - i < lenB) { // If the remaining source string is shorter than key word.
// Means the key word is impossible to be found.
System.out.println("Not found!");
return;
}
// Check the char one by one.
for (int j = 0; j < lenB; j++) {
if (a[i + j] == b[j]) {
if (j == lenB - 1) { // If this char is the last one of key word, means it's found!
System.out.println("Found!");
return;
}
} else {
// If any char mismatch, then right shift 1 char in the source string and restart the search
break;
}
}
}
}
}
You can just use String.contains();
If you really want to implement a method, try this one:
public static void main(String[] args) {
// Initialize values searchedWord and original by user
String original = [get original word from user] ;
String searchedWord = [get searched for word from user];
boolean containsWord = false;
int comparePosition = 0;
for(int i = 0; i < original.length() - searchedWord.length(); i++) {
if(original.charAt(i) == searchedWord.charAt(comparePosition)) {
comparePosition += 1;
} else {
comparePosition = 0;
}
if(comparePosition == searchedWord.length()) {
containsWord = true;
break;
}
}
return containsWord? "Word found!!" : "Word not found.";
}