I have only 6 months of Java experience (and I'm also new here) so please bear with me if things don't look entirely right in my code. Please note that it's still a work in progress. I'm trying to write a program that takes in strings and prints only the ones that are palindromes.
I'm supposed to:
- create a method named isPalindrome, which has a String parameter and
- returns a Boolean based on whether the string is a palindrome or not. Then
- modify the main method to use isPalindrome to print only the palindromes.
For example, if I type: "madam James apple mom timer", it should print "madam" and "mom".
This is basically the program I am trying to write:
Ex: Let's use the word "madam". The program will check if the first and last letters match ("madam"). If that is true, then it'll check the next letters, this time "a" and "a" ("madam). And so on and so forth.
This is the Java code I have so far:
public class Palindrome
{
private String theWord; //Error: The value of the field Palindrome.theWord is not used
public boolean isPalindrome( String theWord ) {
int firstPointer = 0;
int secondPointer = theWord.length() - 1;
for ( int i = 0; i < theWord.length( ); i++ ) {
if ( theWord.charAt[0] == theWord.charAt (theWord.length() - 1) ) { //Error: charAt cannot be resolved or is not a field
return true;
}
return false;
}
}
public static void main( String[] theWord ) {
Palindrome = new Palindrome( ); //Error: Palindrome cannot be resolved to a variable
for ( int i = 0; i < theWord.length; i++ ) {
while (firstPointer < secondPointer) { //Error: "firstPointer" cannot be resolved to a variable. "secondPointer" cannot be resolved to a variable
if ( theWord.charAt[0] == theWord.charAt (theWord.length() - 1) ) { //Error: charAt cannot be resolved to a variable or is not a field. Cannot invoke length() on the array type String[]
firstPointer++; //Error: "firstPointer" cannot be resolved to a variable
secondPointer++; //Error: "secondPointer" cannot be resolved to a variable
}
System.out.println(theWord);
}
}
}
}
Any bit of help knowing where I've gone wrong would be greatly appreciated. Please don't just give me the right code. I would like to figure this out. Thank you very much.
**EDIT: I've included the errors as comments in the code now. I'm using Eclipse by the way.
-->**EDIT 2: Okay guys. I've read most of your answers and have been able to correct most of my code so far (Thank you all so much so far). The only part I'm still having an issue with right now is this part:
if ( theWord.charAt(i) == theWord.charAt (theWord.length() - i - 1) ) {
leftPointer++;
rightPointer--;
I'm now getting a "Cannot invoke charAt(int) on the array type String[]"
and "Cannot invoke length() on the array type String[]".
Those are the only two errors remaining, then I'll test the code out. I've been trying to resolve them for a while now but I'm still not entirely sure what those errors mean.
Eclipse is suggesting that I change theWord.charAt(i) to theWord.length which is not what I want. It is also suggesting I remove "( )" from length but I don't think that's right either.
Looking at your isPalindrome method :
if ( theWord.charAt(0) == theWord.charAt (theWord.length() - 1)
here you always compare the first character to the last character. In each iteration you should compare a different pair of characters, until you find a pair that doesn't match, or reach the middle of the word.
You should use the i variable of your loop :
if ( theWord.charAt(i) == theWord.charAt (theWord.length() - i - 1)
And the return value should be the exact opposite. If you find a pair of characters that don't match, you return false. Only if the loop ends without returning false, you return true.
Okay, let's break everything down into little sizable chunks.
Input a string
Parse the string, check if it is a palindrome.
Print out the words in the string which were palindromes.
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter a sentence: ");
String sentence = scan.nextLine(); // 1.
String[] words = sentence.split(" ");
for (String word : words) { // 3.
if (isPalindrome(word)) {
System.out.println(word);
}
}
}
/**
* Check if the string is a palindrome.
* #param string
* #return True if string is palindrome.
*/
public static boolean isPalindrome(String string) { // 2.
for (int i = 0; i < string.length() / 2; i++) {
if (string.charAt(i) != string.charAt(string.length() - i - 1)) {
return false;
}
}
return true;
}}
Some explanation
The method/function isPalindrome is static because we are calling it from a static context, that is the main function. If you want to use it non-statically you would place it in a class and create a object from that class. The rest should be understandable. :-)
A better isPalindrome method
The shortest way is probably just following the definition:
If you reverse the string, and it's still the same, then it's a palindrome:
public static boolean isPalindrome(String input)
{
String reverse = new StringBuilder(input).reverse().toString();
return input.equalsIgnoreCase(reverse);
}
But if there is an educational goal (?), and iterators should be used for some reason then, imho it makes more sense to iterate from the outside towards the inside of the string.
public static boolean isPalindrome(String input)
{
int length = input.length();
for (int i = 0; i < length/2 ; i++)
{
if (input.charAt(i) != (input.charAt(length-1-i))) return false;
}
return true;
}
Phrase parsing
In your example you used the input of the main String[] parameter. Here is just some information in case you wanted to split it to words manually.
Equivalent to what you got now:
String[] words = phrase.split("\\s+");
for (String word : words)
{
// do stuff
}
The split method uses a delimiter to split a String into a String[]. The delimiter \\s is a regex that represents all kinds of whitespace (not only spaces, but also tabs, new-line characters, etc ...).
But it's not perfect (and neither is your way), there can still be commas, dots and other marks in the phrase. You could filter these characters in an iteration, using the Character.isLetterOrDigit method. Alternatively, you could just perform a replace(...) to remove comma's, points and other marks. Or you could use more complex regular expressions as well.
About your code
The first error message : "The value of the field is not used".
The error message is caused by the global private field theWord, because it is never used. It's not used because you also have a parameter with the same name inside the method isPalindrom(String theWord). Whenever you reference theWord inside that method, it will always give advantage to method arguments before considering global variables.
It looks like you are stuck here with a design contradiction.
What exactly is the class Palindrome ? There are 2 options:
Is it supposed to be a toolbox like the Math class ? like boolean value = Palindrome.isPalindrome("madam");?
Or is it supposed to be an Object that you instantiate using a constructor ? like boolean value = new Palindrome("madam").isPalindrome();
Option 1: a toolbox:
public class Palindrome
{
// removed the private field theWord
// make this method static !!
public static boolean isPalindrome( String theWord ) {
...
}
public static void main( String[] theWord ) {
// remove the Palindrome object
// inside the loop check use the static method
// which does not require an object.
if ( Palindrome.isPalindrome(word))
{
}
}
}
Option 2: an object
public class Palindrome
{
// keep the private field theWord
private String theWord;
public Palindrome(String theWord)
{
// set the value of the argument to the private field
this.theWord = theWord;
}
// don't make this method static
// also you don't need the parameter any more.
// it will now use the global field theWord instead of a parameter.
public boolean isPalindrome() {
...
}
public static void main( String[] theWord ) {
// inside the loop check use an object
Palindrome palindrome = new Palindrome(word);
if ( palindrome.isPalindrome())
{
}
}
As for the errors about the firstPointer and secondPointer. You need to define and initialize those variables. I.e. put int firstPointer = 0; before the loop.
In the loop check it out this way:
boolean isPalin = true;
for ( int i = 0; i < theWord.length( )/2; i++ ) { // loop goes till half since no need to check after that
if ( !(theWord.charAt(i) == theWord.charAt (theWord.length() - 1 - i)) ) { // will check each letter with each end letter
isPalin = false;
break;
}
}
return isPalin;
Another things to add -
1 -firstPointer secondPointer are local variables to isPalindrome
2 - When u have decalared theWord as global variable there doent seems a need to pass it. You can use it within the same class.
3 - theWord in main(String[] theWord) would require you to provide input as arguments, it better you go for console input at runtime.
4 - In main you should split each word and pass it to isPalindrome. In your code you are not calling isPalindrome to check anywhere.
Related
A class words defines a recursive function to perform string related operations. The class details
are given below:
Class name : words
Data members/instance variables
text : to store string.
w : integer variable to store total words.
Member functions/methods
words( ) : constructor to store blank to string and 0 to integer data.
void Accept( ) : to read a sentence in text. Note that the sentence may contain more
than one blank space between words.
int FindWords(int) : to count total number of words present in text using Recursive
Technique and store in ‘w’ and return.
void Result( ) : to display the original string. Print total number of words stored in
‘w’ by invoking the recursive function.
I tried this code
public static int CountWords(String str) {
int c = 0;
int i = str.indexOf(" ");
if (str.isEmpty()) {
return 0;
}else
if (i == str.indexOf(" ")) {
return c++;
}
//str.substring(0,str.indexOf(" ")-1);
c++;
return c + CountWords(str.substring(i + 1));
}
but i need to return an integer value and i am confused with that..
In your code, the last return statement is inaccessible. Reason: you have put an if-else block and have put return in both the cases. So the function actually gets returned from the if-else block itself (within else, the condition of if is always true since you assigned the very value i.e. str.indexOf(" ")).
I have written down the code according to the question you gave above...
public int findWords(int i){
if(i > text.lastIndexOf(" "))
return 1;
i = text.substring(i).indexOf(" ") + i;
if(i < 0)
return 1;
if(text.substring(i).equals(null))
return 0;
return( findWords(i+1) + 1);
}
Hope you find it well working.
Your function already is returning a integer, it just happens to always be 0.
This is due to
else if (i == str.indexOf(" ")) {
return c++;
}
Always being true and c++ only updating after the return statement was passed.
This happens because you already set i to be the indexOf(" ") and due to the implementation of incrementation using int++. Also, keep in mind hat you need to increase the number of words by 2 here, since you're ending the function between two words.
Therefore, use this instead:
else if (i == str.lastIndexOf(" ")) {
return c+2;
}
You should see that now the function is returning the correct amount of words.
This is a assignment I'm doing and it seems I can't get it to work properly.
The question is below.
A palindrome is a word or phrase that reads the same forward and
backward, ignoring blanks and considering uppercase and lowercase
versions of the same letter to be equal.for example,the following are
palindromes:
warts n straw
radar
able was I ere I saw Elba
xyzczyx
Write a program that will accept a sequence of characters terminated
by a period and will decide whether the string--without the
period---is a palindrome.You may assume that the input contains only
letters and blanks and is at most 80 characters long.Include a loop
that allows the user to check additional strings until she or he
requests that the program end.
Hint: Define a static method called isPalindrome that begins as
follows:
Precondition: The array a contains letters and blanks in
positions a[0] through a[used - 1]. Returns true if the string is a
palindrome and false otherwise.
public static boolean isPalindrome(char[] a, int used)
Your program should read the input characters into an array whose base
type is char and then call the preceding method. The int variable used
keeps track of how much of the array is used, as described in the
section entitled "Partially Filled Arrays."
This is my class code:
public class Palindrome_class
{
// instance variable
char[] characterArray;
//constructor
//#param data is a string of characters
public Palindrome_class(String data)
{
characterArray = data.toUpperCase().toCharArray();
}
//#return true if the word is a palindrome, otherwise returns false.
public boolean isPalindrome(char[] a, int used)
{
int i = 0, j = used - 1;
while (i < j)
{
if(characterArray[i] == characterArray[j])
{
i++;
j--;
}
else
{
return false;
}
}
return true;
}
}
This is my main code:
import java.util.Scanner;
public class palindromeTest
{
public static void main(String[] args)
{
int used = 0;
char[] chars = new char[80];
Scanner inputWord = new Scanner(System.in);
Scanner reply = new Scanner(System.in);
System.out.println("Enter a string characters, terminated by a period.");
String data;
String cq;
Palindrome_class word;
do
{
//input word from user.
data = inputWord.nextLine();
word = new Palindrome_class(data);
//check for palindrome.
if(word.isPalindrome(chars, used))
System.out.println(data + " is a palindrome.");
else
System.out.println(data + " is not a palindrome.");
//request to continue or quit.
System.out.println("Continue or Quit?");
cq = reply.nextLine();
}
while (cq.equalsIgnoreCase("continue"));
System.exit(0);
}
}
This is the results:
Enter a string characters, terminated by a period.
radar.
radar. is a palindrome.
Continue or Quit?
continue
use
use is a palindrome.
Continue or Quit?
continue
use.
use. is a palindrome.
Continue or Quit?
continue
apple.
apple. is a palindrome.
Continue or Quit?
Quit
Please tell me where I'm making a mistake.
You are checking whether a String is a palindrome with this call :
if(word.isPalindrome(chars, used))
However, used is 0, so your method always returns true.
You are also ignoring the instructions of your assignment. You are not doing anything with the chars array, you are not removing the period that's supposed to be at the end of the input String, your isPalindrome method is not static, etc...
U did a very little mistake.. U are sending "used" variable as 0 each time. ideally it should be length of a word.
please check it. use
used = data.length();
before sending it to the check method
This is from cracking the Coding Interview Book
The Questions Implement an algorithm to determine if a string has all unique characters. What if
you can not use additional data structures?
I am wondering what is happening in the if statement below? can anyone explain it to me ?
I have left my understanding of the code in the comments.Please correct me if i am wrong
public class Uniquechar2 {
public static boolean isUniqueChars2(String str) {
// Create a new boolean array of 256 characters to account for basic a cii and extended ascii characters
boolean[] charSet = new boolean[256];
//iterate through the array
for (int i = 0; i < str.length(); i++) {
// Assign the value of current value of the iterator i to int variable val.So if we are looping through "hello" at i = 0 the int value of 'h' will be assigned to val.Is that correct?
int val = str.charAt(i);
// Continuing from the example of loping throughout the string "hello" the if statement will see if 'h' is in charSet and since it will be there it will return false /is that what is happening?
if (charSet[val]) {
return false;
}
// Is this the else statement? true will be assigned to charSet[h] in this case
charSet[val] = true;
}
// I dont understand why we are returning true at the end ?
return true;
}
public static boolean isUniqueChars2(String str) {
// Create a new boolean array of 256 characters to account for basic ascii and extended ascii characters
boolean[] char_set = new boolean[256];
// Iterate through the string we are testing
for (int i = 0; i < str.length(); i++) {
// Get the numerical (ascii) value of the character in the `str` at position `i`.
int val = str.charAt(i);
// If char_set[val] has been set, that means that this character was already present in the string. (so in string 'hello' this would be true for the second 'l')
if (char_set[val]) {
return false;
}
// If the character hasn't been encountered yet (otherwise we would have returned false above), then mark this particular character as present in the string
char_set[val] = true;
}
// If the function hasn't returned false after going through the entire string that means that each character is unique - thus returning true
return true;
}
Is this the else statement
No, otherwise there would be an else in the code. But in this case, else is unnecessary since, if char_set[val] is true, the execution of the method stops immediately, due to the return false; instruction.
I dont understand why we are returning true at the end ?
Because since no duplicate has been found, the method must return true to indicate that the string is composed of unique characters. If a duplicate had been found, the method would have returned already in
if (char_set[val]) {
return false;
}
I would just use regex, which requires only one line of code:
public static boolean isUniqueChars(String str) {
return str.matches("((.)(?!.*?\\2))*");
}
Breaking down the regex:
(.) captures every character
(?!.*?\\2) is a negative look ahead for a back reference to the captured group
Together, these mean "a character that does not reappear after itself"
(...)* around the above means 0-n of them
Altogether, it means "comprised of characters that do do reappear later in the string", ie unique characters.
One-line solution without any extra data structure:
str.chars().distinct().count() == (int)str.length();
I have to be able to input any two words as a string. Invoke a method that takes that string and returns the first word. Lastly display that word.
The method has to be a for loop method. I kind of know how to use substring, and I know how to return the first word by just using .substring(0,x) x being how long the first word is.
How can I make it so that no matter what phrase I use for the string, it will always return the first word? And please explain what you do, because this is my first year in a CS class. Thank you!
I have to be able to input any two words as a string
The zero, one, infinity design rule says there is no such thing as two. Lets design it to work with any number of words.
String words = "One two many lots"; // This will be our input
and then invoke and display the first word returned from the method,
So we need a method that takes a String and returns a String.
// Method that returns the first word
public static String firstWord(String input) {
return input.split(" ")[0]; // Create array of words and return the 0th word
}
static lets us call it from main without needing to create instances of anything. public lets us call it from another class if we want.
.split(" ") creates an array of Strings delimited at every space.
[0] indexes into that array and gives the first word since arrays in java are zero indexed (they start counting at 0).
and the method has to be a for loop method
Ah crap, then we have to do it the hard way.
// Method that returns the first word
public static String firstWord(String input) {
String result = ""; // Return empty string if no space found
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
result = input.substring(0, i);
break; // because we're done
}
}
return result;
}
I kind of know how to use substring, and I know how to return the first word by just using .substring(0,x) x being how long the first word is.
There it is, using those methods you mentioned and the for loop. What more could you want?
But how can I make it so that no matter what phrase I use for the string, it will always return the first word?
Man you're picky :) OK fine:
// Method that returns the first word
public static String firstWord(String input) {
String result = input; // if no space found later, input is the first word
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
result = input.substring(0, i);
break;
}
}
return result;
}
Put it all together it looks like this:
public class FirstWord {
public static void main(String[] args) throws Exception
{
String words = "One two many lots"; // This will be our input
System.out.println(firstWord(words));
}
// Method that returns the first word
public static String firstWord(String input) {
for(int i = 0; i < input.length(); i++)
{
if(input.charAt(i) == ' ')
{
return input.substring(0, i);
}
}
return input;
}
}
And it prints this:
One
Hey wait, you changed the firstWord method there.
Yeah I did. This style avoids the need for a result string. Multiple returns are frowned on by old programmers that never got used to garbage collected languages or using finally. They want one place to clean up their resources but this is java so we don't care. Which style you should use depends on your instructor.
And please explain what you do, because this is my first year in a CS class. Thank you!
What do I do? I post awesome! :)
Hope it helps.
String line = "Hello my name is...";
int spaceIndex = line.indexOf(" ");
String firstWord = line.subString(0, spaceIndex);
So, you can think of line as an array of chars. Therefore, line.indexOf(" ") gets the index of the space in the line variable. Then, the substring part uses that information to get all of the characters leading up to spaceIndex. So, if space index is 5, it will the substring method will return the indexes of 0,1,2,3,4. This is therefore going to return your first word.
The first word is probably the substring that comes before the first space. So write:
int x = input.indexOf(" ");
But what if there is no space? x will be equal to -1, so you'll need to adjust it to the very end of the input:
if (x==-1) { x = input.length(); }
Then use that in your substring method, just as you were planning. Now you just have to handle the case where input is the blank string "", since there is no first word in that case.
Since you did not specify the order and what you consider as a word, I'll assume that you want to check in given sentence, until the first space.
Simply do
int indexOfSpace = sentence.indexOf(" ");
firstWord = indexOfSpace == -1 ? sentence : sentence.substring(0, indexOfSpace);
Note that this will give an IndexOutOfBoundException if there is no space in the sentence.
An alternative would be
String sentences[] = sentence.split(" ");
String firstWord = sentence[0];
Of if you really need a loop,
String firstWord = sentence;
for(int i = 0; i < sentence.length(); i++)
{
if(sentence.charAt(i) == ' ')
{
sentence = firstWord.substring(0, i);
break;
}
}
You may get the position of the 'space' character in the input string using String.indexOf(String str) which returns the index of the first occurrence of the string in passed to the method.
E.g.:
int spaceIndex = input.indexOf(" ");
String firstWord = input.substring(0, spaceIndex);
Maybe this can help you figure out the solution to your problem. Most users on this site don't like doing homework for students, before you ask a question, make sure to go over your ISC book examples. They're really helpful.
String Str = new String("Welcome to Stackoverflow");
System.out.print("Return Value :" );
System.out.println(Str.substring(5) );
System.out.print("Return Value :" );
System.out.println(Str.substring(5, 10) );
I have a string: LOAN,NEFT,TRAN. I want to substring the string based on getting a , during traversing the string. So I tried to first get a count for how many , are there. but not sure what function to user to get what I want. Also this should be dynamic, meaning I should be able to create as many substrings as required based on number of ,s. I tried the following code:
package try1;
public class StringTest {
public static void main(String[] args) {
String str="LOAN,NEFT,TRAN";
int strlen=str.length();
int count=0;
for(int i=0;i<strlen;i++)
{
if(str.contains("'"))
count++;
}
System.out.println(""+count);
for (int j=0;j<count;j++)
{
//code to create multiple substrings out of str
}
}
}
But I do not think contains() is the function I am looking for because value of count here is coming 0. What should I use?
Your code doesn't actually count the , characters because 1) contains doesn't take into account your loop variable 2) it's searching for ', not ,
Assuming you want to work at a low level rather than using high level functions like .split(), then I'd recommend the following.
for(char c : str.toCharArray()) {
if (c == ',') {
count++;
}
}
You can use split to get substrings directly:
String[] substrings = str.split(",");
Is this what you want as an output: (shown below)?
["LOAN", "NEFT", "TRAN"] // Array as an output
Or to just get the count of the splitting char, you can use the same line as above with this:
int count = substrings.length - 1;