I'm trying to insert in a for loop to a Stringbuilder but it crashes. This is my code
StringBuilder sb = new StringBuilder("This is a text");
for(int i = 0;i < sb.length();i++){
if(sb.charAt(i) == 't'){
sb.insert(i, 't');
}
}
Purpose of this is to double every 't'.
You are getting an OutOfMemoryError, because you don't skip the t character that you're doubling.
This is a text
^
This is a ttext
^
This is a tttext
^
This continues until you've run out of memory.
You must skip the t that you just added by incrementing i just after inserting the doubled t.
if (sb.charAt(i) == 't')
{
sb.insert(i, 't');
i++;
}
I suggest to use a StringBuilder and the string like this:
String currentText = "This is a text";
StringBuilder sb = new StringBuilder();
for(int i = 0; i < currentText.length(); i++){
sb.append(currentText.charAt(i));
if(currentText.charAt(i) == 't'){
sb.append('t');
}
}
With that solution, you will not have the problem with a never ending loop.
The problem you are having is the following.
Iterate through the list, from start to finish.
If you find a t add a t.
Two scenarios, either there is at least 1 t in the String or there are none. If there are none then all will work, the code does nothing. However if there is a t it will find that t add another t, the next character to check will now be... a t!. Infinite loop.
You can go around the problem by either by incrementing i (ì++) or by using continue.
Because when you find a t this instruction sb.insert(i, 't') add t in position i and previous t become in position i+1, so when i will increment in the end of for loop with i++ you deal another time with same letter t. so that will cause infinity loop.
To fix that you should increment i with i++ in if block like that:
if(sb.charAt(i) == 't'){
sb.insert(i, 't');
i++;
}
Related
So I'm trying to solve the Longest Substring Without Repeating Character problem in a webpage and when I'm trying to upload it it will show me this bug:
class Solution {
public int lengthOfLongestSubstring(String s) {
HashSet<Character> hash = new HashSet<>();
int count = 0, finalCount = 1;
char prevChar = s.charAt(0);
hash.add(prevChar);
for (int i = 1; i < s.length(); i++)
{
char character = s.charAt(i);
if (!hash.contains(character)){
hash.add(character);
count++;
if (count > finalCount) finalCount = count;
}
else{
hash.clear();
hash.add(character);
count = 1;
}
prevChar = character;
}
return finalCount;
} }
Is there anything wrong with it?
If not, do you think my algorithm was efficient? I can't compare its performance since the webpage won't let me upload it.
You call s.charAt(0) in line 5. I imagine they pass in the empty string as a test case and you are getting an out of bounds exception. Prior to line 5 add a check to see if the string length is 0 and if it is return 0.
According to the error description it's doing a dummy-spit at line 5 of the Solution class.
Based on the picture that's:
char prevChar = s.charAt(0);
The error is ArrayIndexOutOfBounds which generally indicates you tried to get more out of something than was actually there (e.g. running over the end of an array).
Here I'd suggest maybe putting in some System.out.println lines at line 3 to sanity check the method parameter, e.g.:
(a) if the input String s is null
or
(b) if the input String s is empty (e.g. "")
charAt(0) will get the first character, but if there are zero characters then trying to get the 1th character is an error, no?
NB: something like this:
System.out.println("Input was :" + s + ":");
Will show both of those conditions, as either:
Input was ::
for an empty String
Input was :null:
for a null String
I need to create a loop that adds "o" after each consonant
I am going to walk you through what I corrected and changed in your code to make it work in order to make it quick and easy for you to comprehend why your code doesn't work and why my answer fixes it.
The mistakes you made are basic ones and frankly you shouldn't have to much of a hard time correcting them yourself if you would use a debugger that walks you step by step in how your code works. You should look on how to use a debugger (for example the debugger used in Eclipse, hopefully you are using an IDE to make your life easier).
Firstly, when you are looking for a consonant in your code, you are only walking through the half of it because of your condition for(int x = 0; x<20; x++) since your string holding the consonants if of a length of 40 characters. This means you are missing consonants like the letter s.
Then you are correctly the consonants you find according to your Swedish language game. But you are never handling characters that are not of these found consonants. You should make a case where you handle these "non consonant" letters, may they be vowels or any kind of character (like punctuation marks and so on). I am fixing this with the use of a simple boolean here.
Keep in mind that my goal here is to change your code as little as I can, thus I went for adding a boolean to handle your cases (checking the presence of a consonant). There are, obviously, many other ways to implement what you are trying to do.
Here come the changes you should add to your code:
/*This comes after your print "På rövarspråk:"*/
boolean isConsonant = false; //Boolean to check wether there is a consonant or not
for(int i = 0; i<length; i++) {
//You didn't go through the whole consonants list you made with your prevision condition
for(int x = 0; x<consonants.length; x++){
if(array[i] == consonants[x])
{
isConsonant = true; //Set the boolean accordingly
String add = array[i]+"o"+array[i];
slang = slang + add;
break;
}
}
if(!isConsonant){ //If we don't have a consonant, add the char to the result string
slang += array[i];
}
isConsonant = false; //Reset the boolean for the next character
}
/*Here you can print the result (slang) if you want, as you did*/
so the idea is to dublicate consonants and put "o" between them, like t becomes tot, s becomes sos. Vocals are just copied. So you need a method that tells you if a given character is a vocal or consonant to base your decision on that.
public static boolean isConsonant(char inputChar){
final String consonantsx = "bBcCdDfFgGhHjJkKlLmMnNpPqQrRsStTvVwWxXzZ";
char consonants[] = consonantsx.toCharArray(); // String to charr
for(int i=0; i < consonants.length;i++){
if(inputChar == consonants[i]){ //note that in Strings u use the equals method instead of "=="
return true;
}
}
return false;
}
Given this method you can use it in the "translator method".
public String rovarSpraket(String normalString) {
char[] array = normalString.toCharArray(); // Input to a char array
System.out.println("På rövarspråk:");
String slang = "";
for (int i = 0; i < normalString.length(); i++) {
String add = "" + array[i];
if(Goran.isConsonant(array[i])){
add += "o" + array[i];
}
slang += add;
}
return slang;
}
This translates stubborn to sostotubobboborornon like in the wikipedia article https://en.wikipedia.org/wiki/R%C3%B6varspr%C3%A5ket.
I am writing a program that is going to read a string from a file, and then remove anything that isn't 1-9 or A-Z or a-z. The A-Z values need to become lowercase. Everything seems to run fine, I have no errors, however my output is messed up. It seems to skip certain characters for no reason whatsoever. I've looked at it and tweaked it but nothing works. Can't figure out why it is randomly skipping certain characters because I believe my if statements are correct. Here is the code:
String dataIn;
int temp;
String newstring= "";
BufferedReader file = new BufferedReader(new FileReader("palDataIn.txt"));
while((dataIn=file.readLine())!=null)
{
newstring="";
for(int i=0;i<dataIn.length();i++)
{
temp=(int)dataIn.charAt(i);
if(temp>46&&temp<58)
{
newstring=newstring+dataIn.charAt(i);
}
if(temp>96&&temp<123)
{
newstring=newstring+dataIn.charAt(i);
}
if(temp>64&&temp<91)
{
newstring=newstring+Character.toLowerCase(dataIn.charAt(i));
}
i++;
}
System.out.println(newstring);
}
So to give you an example, the first string I read in is :
A sample line this is.
The output after my program runs through it is this:
asmlietis
So it is reading the A making it lowercase, skips the space like it is suppose to, reads the s in, but then for some reason skips the "a" and the "m" and goes to the "p".
You're incrementing i in the each of the blocks as well as in the main loop "header". Indeed, because you've got one i++; in an else statement for the last if statement, you're sometimes incrementing i twice during the loop.
Just get rid of all the i++; statements other than the one in the for statement declaration. For example:
newstring="";
for(int i=0;i<dataIn.length();i++)
{
temp=(int)dataIn.charAt(i);
if(temp>46&&temp<58)
{
newstring=newstring+dataIn.charAt(i);
}
if(temp>96&&temp<123)
{
newstring=newstring+dataIn.charAt(i);
}
if(temp>64&&temp<91)
{
newstring=newstring+Character.toLowerCase(dataIn.charAt(i));
}
}
I wouldn't stop editing there though. I'd also:
Use a char instead of an int as the local variable for the current character you're looking at
Use character literals for comparisons, to make it much clearer what's going on
Use a StringBuilder to build up the string
Declare the variable for the output string for the current line within the loop
Use if / else if to make it clear you're only expecting to go into one branch
Combine the two paths that both append the character as-is
Fix the condition for numbers (it's incorrect at the moment)
Use more whitespace for clarity
Specify a locale in toLower to avoid "the Turkey problem" with I
So:
String line;
while((line = file.readLine()) != null)
{
StringBuilder builder = new StringBuilder(line.length());
for (int i = 0; i < line.length(); i++) {
char current = line.charAt(i);
// Are you sure you want to trim 0?
if ((current >= '1' && current <= '9') ||
(current >= 'a' && current <= 'z')) {
builder.append(current);
} else if (current >= 'A' && current <= 'Z') {
builder.append(Character.toLowerCase(current, Locale.US));
}
}
System.out.println(builder);
}
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);}
}
I am tasked to create a function that checks if a number of small "(" and big "[" parentheses and closes properly. For example...
(()) [] ([])[(())]
...is correct, but...
() [(])
...is incorrect.
Any advice on how to start approaching this problem? The use of recursive functions is prohibited.
This is what I have so far:
{
int nr_ap_x = 0; // fie x - "("
int nr_ap_y = 0; // fie y - ")"
boolean corect = true;
for (int i=0; i < sir.length; i++)
{
if (sir[i].compareTo("(") == 0) nr_ap_x++;
else
if (sir[i].compareTo(")") == 0) nr_ap_y++;
if (nr_ap_x < nr_ap_y) corect = false;
}
if (nr_ap_x != nr_ap_y) corect = false;
if (corect) System.out.println("Parantezele sunt inchise corect ! ");
else System.out.println("Parantezele NU sunt inchise corect ! ");
}
General outline of how to do this (assuming you have an array of strings each containing a parenthesis):
Create a stack which contains strings.
For each "(" or "[", push onto the stack.
For each ")" or "]", check if the top of the stack matches the type of parenthesis.
If it does, discard the top of the stack.
Otherwise, the parenthesis are mismatched.
If you have a non empty stack once all your strings run out, you also have mismatched parentheses.
Hint: keep a data structure containing the opening glyphs, and use and update it when a closing glyph appear.
Anything that's done with recursion can also be done with iteration. If you already know how to do it recursively then map it to a while loop. This is explained in Replace Recursion with Iteration.
You could use a deterministic finite automaton - http://en.wikipedia.org/wiki/Deterministic_finite-state_machine