I'm working on a program for Java on how to find a list of palindromes that are embedded in a word list file. I'm in an intro to Java class so any sort of help or guidance will be greatly appreciated!
Here is the code I have so far:
import java.util.Scanner;
import java.io.File;
class Palindromes {
public static void main(String[] args) throws Exception {
String pathname = "/users/abrick/resources/american-english-insane";
File dictionary = new File(pathname);
Scanner reader = new Scanner(dictionary);
while (reader.hasNext()) {
String word = reader.nextLine();
for (int i = 0; i > word.length(); i++) {
if (word.charAt(word.indexOf(i) == word.charAt(word.indexOf(i)) - 1) {
System.out.println(word);
}
}
}
}
}
There are 3 words that are 7 letters or longer in the list that I am importing.
You have a few ways to solve this problem.
A word is considered a palindrome if:
It can be read the same way backwards as forwards.
The first element is the same as the last element, up until we reach the middle.
Half of the word is the same as the other half, reversed.
A word of length 1 is trivially a palindrome.
Ultimately, your method isn't doing much of that. In fact, you're not doing any validation at all - you're only printing the word if the first and last character match.
Here's a proposal: Let's read each end of the String, and see if it's a palindrome. We have to take into account the case that it could potentially be empty, or be of length 1. We also want to get rid of any white space in the string, as that can cause errors on validation - we use replaceAll("\\s", "") to solve that.
public boolean isPalindrome(String theString) {
if(theString.length() == 0) {
throw new IllegalStateException("I wouldn't expect a word to be zero-length");
}
if(theString.length() == 1) {
return true;
} else {
char[] wordArr = theString.replaceAll("\\s", "").toLowerCase().toCharArray();
for(int i = 0, j = wordArr.length - 1; i < wordArr.length / 2; i++, j--) {
if(wordArr[i] != wordArr[j]) {
return false;
}
}
return true;
}
}
I'm assuming that you're reading in strings. Use string.toCharArray() to convert each string to a char[]. Iterate through the character array using a for loop as follows: on iteration 1, if the first character is equal to the last character, then proceed to the next iteration, else return false. On iteration 2, if the second character is equal to the second-to-last character then proceed to the next iteration, else return false. And so on, until you reach the middle of the string, at which point you return true. Be careful of off-by-one errors; some strings will have an even length, some will have an odd length.
If your palindrome checker is case insensitive, then use string.toLowerCase().toCharArray() to preprocess the character array.
You can use string.charAt(i) instead of string.toCharArray() in the for loop; in this case, if the palindrome checker is case insensitive then preprocess the string with string = string.toLowerCase()
Let's break the problem down: In the end, you are checking if the reverse of the word is equal to the word. I'm going to assume you have all of the words stored in an array called wordArray[].
I have some code for getting the reverse of the word (copied from here):
public String reverse(String str) {
if ((null == str) || (str.length() <= 1)) {
return str;
}
return new StringBuffer(str).reverse().toString();
}
So, now we just need to call that on every word. So:
for(int count = 0; count<wordArray.length;count++) {
String currentWord = wordArray[count];
if(currentWord.equals(reverse(currentWord)) {
//it's a palendrome, do something
}
}
Since this is homework, i'll not supply you with code.
When i code, the first thing i do is take a step back and ask myself,
"what am i trying to get the computer to do that i would do myself?"
Ok, so you've got this huuuuge string. Probably something like this: "lkasjdfkajsdf adda aksdjfkasdjf ghhg kajsdfkajsdf oopoo"
etc..
A string's length will either be odd or even. So, first, check that.
The odd/even will be used to figure out how many letters to read in.
If the word is odd, read in ((length-1)/2) characters.
if even (length/2) characters.
Then, compare those characters to the last characters. Notice that you'll need to skip the middle character for an odd-lengthed string.
Instead of what you have above, which checks the 1st and 2nd, then 2nd and 3rd, then 3rd and fourth characters, check from the front and back inwards, like so.
while (reader.hasNext()) {
String word = reader.nextLine();
boolean checker = true;
for (int i = 0; i < word.length(); i++) {
if(word.length()<2){return;}
if (word.charAt(i) != word.charAt(word.length()-i) {
checker = false;
}
}
if(checker == true)
{System.out.println(word);}
}
Related
I am writing a hangman program and one of the requirements to a hangman game is preventing the user from entering the same letter twice.
I have written the code for that, but the problem is every time I enter a letter it says it is already entered. I need to only say it when it is entered the second time. You guys have any suggestions? I've been trying to fix this for the past few hours, I figured I could ask on here to find some help. I already looked at another Stackoverflow question regarding something similar to this but all those solutions have the same result.
In Java, how can I determine if a char array contains a particular character?
I've tried something like this but it won't work either:
boolean contains = false;
for (char c : store) {
if (c == character) {
System.out.println("Already Entered");
contains = true;
break;
}
}
if (contains) {
// loop to top
continue;
}
SECOND CLASS-
public void hangman(String word, int life) {
KeyboardReader reader = new KeyboardReader();
char[] letter = new char[word.length()];
char[] store = new char[word.length()];
String guess;
int i = 0, tries = 0, incorrect = 0, count = 1, v = 0;
while (i < word.length()) {
letter[i] = '-';
I would just use the String.contains() method:
String aString = "abc";
char aChar = 'a';
return aString.contains(aChar + "");
To keep track of guessed letters you can use a StringBuffer, appending them using a StringBuffer.append() to append new letters (maintaining state) and use the StringBuffer.toString() method to get the String representation when you need to do the comparison above.
Since Java 1.5 the class String contains the method contains(). My idea is to collect all entered letters into a string variable and using above method:
// My code line
String letterAlreadyEntered = "";
// Your code line
char character = reader.readLine().charAt(0);
// My code line
if (letterAlreadyEntered.contains("" + character) == true) {
//Put code here what ever you want to do with existing letter
} else {
letterAlreadyEntered += character;
}
In my opinion, this is an easier way to check for occurrences than in arrays, where you have to write your own check method.
the code below is working properly i just can't understand that part in the code
if (temp.indexOf(Search.charAt(j)) == -1){
cpt=0;
} else {
cpt++;
}
what does that condition mean ??
public static void main(String[] args) {
String Search="loop";
String[] words={"loop","pool","lopo","book","kobo","oopl","olop","opol"};
int cpt=0;
String temp;
for(int i=0;i<words.length;i++){
temp=words[i];
for (int j=0 ; j<Search.length();j++) {
if (temp.indexOf(Search.charAt(j))==-1){
cpt=0;
} else {
cpt++;
}
if (cpt==4){
System.out.println("-> :"+ temp);
cpt=0;
}
}
}
}
The idea of the code is to go through each word (call it temp) in the list and compute how many characters in temp are present in Search.
For example "loop" and "pool" both have 1 'p', 1 'l' and 2 'p' characters.
But it's not completely correct...
This line if (temp.indexOf(Search.charAt(j))==-1) is checking to see if Search[j] exists in temp
but it should be more like
if(temp has this Search[j] character at some index (call it t) and t hasn't been used previously)
This should fail for cases like "loop" and "lop"
Also you're hardcoding the number 4 in the code, and it should be replaced by the length of Search
I recommend using a map.
As we know the index of method work on two condition .....
Condition 1 - if the giving character is exist in the string then it means the
character is available at index 0 to string.lenth-1 index.
Condition 2 - if giving character is not exist in string that means the character
is not available in that string from 0 index to entire string,
so is return -1.
I am trying to get my code to prevent a user input from having a number in it.
Essentially I want the code to do as follows:
ask for input
receive input
test whether or not the input contains a number(ex: 5matt vs matt)
if contains a number I want to System.out.println("Error: please do not input a number");
Heres the kicker (and why it's not a duplicate question): I can't use loops or other statements we haven't learned yet. So far the only true statements we've learned are if/else/else if statements. That means I can not use for loops, like some of the answers are suggesting. While they're great answers, and work, I'll lose points for using them.
System.out.println("Please input the first name: ");
String name1 = in.next();
System.out.println("Please input the second name: ");
String name2 = in.next();
System.out.println("Please input the third name: ");
String name3 = in.next();
name1 = name1.substring(0,1).toUpperCase() + name1.substring(1).toLowerCase();
name2 = name2.substring(0,1).toUpperCase() + name2.substring(1).toLowerCase();
name3 = name3.substring(0,1).toUpperCase() + name3.substring(1).toLowerCase();
I have this already but I can't figure out how to test if the input only contains letters.
Okay, there are many ways to deal with this. A good thing would be to use Regex (text matching stuff). But it seems that you should only use very basic comparison methods.
So, let's do something very basic and easy to understand: We iterate over every character of the input and check whether it's a digit or not.
String input = ...
// Iterate over every character
for (int i = 0; i < input.length(); i++) {
char c = s.charAt(i);
// Check whether c is a digit
if (Character.isDigit(c)) {
System.out.println("Do not use digits!");
}
}
This code is very straightforward. But it will continue checking even if a digit was found. You can prevent this using a helper-method and then returning from it:
public boolean containsDigit(String text) {
// Iterate over every character
for (int i = 0; i < input.length(); i++) {
char c = s.charAt(i);
// Check whether c is a digit
if (Character.isDigit(c)) {
return true;
}
}
// Iterated through the text, no digit found
return false;
}
And in your main program you call it this way:
String input = ...
if (containsDigit(input)) {
System.out.println("Do not use digits!");
}
Use a regular expression to filter the input
Eg
str.matches(".*\\d.*")
See this link for more info
There are several ways you could do this, among others:
Iterate over all the chars in the string and check whether any of them is a digit.
Check whether the string contains the number 1, number 2, number 3, etc.
Use a regular expression to check if the string contains a digit.
(Java Regular Expressions)
If you're allowed to define functions, you can essentially use recursion to act as a loop. Probably not what your prof is going for, but you could be just inside the requirements depending on how they're worded.
public static boolean stringHasDigit(String s) {
if (s == null) return false; //null contains no chars
else return stringHasDigit(s, 0);
}
private static boolean stringHasDigit(String s, int index) {
if (index >= s.length()) return false; //reached end of string without finding digit
else if (Character.isDigit(s.charAt(index))) return true; //Found digit
else return stringHasDigit(s, index+1);
}
Only uses if/elseif/else, Character.isDigit, and String.charAt, but recursion might be off limits as well.
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) );
NOTE: For array of counters, one for each UTF-16 character, tallies usage of each char.
For a program I am writing, im trying to tally each character in a specific word. Thus far, i have,
public void tally(String word)
{
if ((word == null) || (word.length() == 0))
return;
(Insert here)
}
For where I have "Insert here" is where i am trying to tally each character in a specific word. Does anyone know simple way of doing it?
Create a map
Start with the 1st char For each character, check if the character key exists in the map, if yes, increase the value of the key by one.
If it doesn't exist, insert the character as key and value as one. when done, print the map.
This look like homework so I'll give some brief pointers
You can split a string into characters with toCharArray
You can iterate over an array with for each
You can store values in an HashMap
public void tally(String word)
{
int count;
if (word == null)
{
return count;
}
else
{
char[] ca;
ca = word.toCharArray();
for(int i = 0; i < ca.length; i++)
{
count++;
}
return count;
}
This will show you how many letters are in the word in total. I think, from this, you can do the rest.