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.
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.
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.
Hi im having this assignment that I don't really understand how to pull off.
Ive been programing java for 2.5 weeks so Im really new.
Im supposed to import a text document into my program and then do these operations, count letters, sentences and average length of words. I've to perform the counting task letter by letter, I'm not allowed to scan the entire document at the same time. Ive managed to import the text and also print it out, but my problem is I cant use my string "line" to do any of these operations. Ive tried converting it to arrays, strings and after a lot of failed attempts im giving up. So how do I convert my input to something I can use, because i always get the error message "line is not a variable" or smth like that.
Jesper
UPDATE WITH MY SOLUTION! also some of it is in Swedish, sorry for that.
Somehow the Format is wrong so I uploaded the code here instead, really don't feel to argue with this wright now!
http://txs.io/3eIb
To count letters, check each character. If it's a space or punctuation, ignore it. Otherwise, it's a letter and we should this increment.
Every word should have a space after it unless it is the last word of the sentence. To get the number of words, track the number of spaces + number of sentences. To get number of sentences, find the number of ! ? and .
I would do that by looking at the ascii value of each character.
int numSentences = 0;
int numWords = 0;
while (line = ...){
for(int i = 0; i <line.length(); i++){
int curCharAsc = (int)(line.at(i)) //get ascii value by casting char to int
if((curCharAsc >= 65 && curCharAsc <= 90) || (curCharAsc >= 97 && curCharAsc <= 122) //check if letter is uppercase or lowercase
numLetters++;
if(curCharAsc == 32){ //ascii for space
numWords++;
}
else if (curCharAsc == 33 || curCharAsc == 46 || curCharAsc == 63){
numWords++;
numSentences++;
}
}
}
double avgWordLength = ((double)(letters))/numWords; //cast to double before dividing to avoid round-off
Your code as presented works fine, it loads a file and prints out the contents line by line. What you probably need to do is capture each of those lines. Java has two useful classes for this StringBuilder or StringBuffer (pick one).
BufferedReader input = new BufferedReader(new FileReader(args[0]));
String line;
StringBuffer buffer = new StringBuffer();
while ((line = input.readLine()) != null) {
System.out.println(line);
buffer.append(line+" ");
}
input.close();
performOperations(buffer.toString());
The only other possibility is (if your own code is not running for you) - possibly you aren't passing the input file name as a parameter when you run this class?
UPDATE
NB - I've modified the line
buffer.append(line+"\n");
to add a space instead of a line break, so that it is compatible with algorithms in the #faraza answer
The method performOperations doesn't exist yet. So you should / could add something like this
public static void performOperations(String data){
}
You method could in turn make calls out to separate methods for each operation
public static void performOperations(String data){
countWords(data);
countLetters(data);
averageWordLength(data);
}
To take it to the next level, and introduce Object Orientation, you could create a class TextStatsCollector.
public class TextStatsCollector{
private final String data;
public TextStatsCollector(final String data) {
this.data = data;
}
public int countWords(){
//word count impl here
}
public int countLetters(){
//letter count impl here
}
public int averageWordLength(){
//average word length impl here
}
public void performOperations(){
System.out.println("Number of Words is " + countWords());
System.out.println("Number of Letters is " + countLetters());
System.out.println("Average word length is " + averageWordLength());
}
}
Then you could use TextStatsCollector like the following in your main method
new TextStatsCollector(buffer.toString()).performOperations();
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;
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);}
}