I have a basic String variable that contains the letter x a total of three times.
I have attempted to find x within the String using charAt, and then print the char and the next two characters next to it.
I have hit a snag within my code and would appreciate any help.
Here is my code.
public class StringX{
public static void main(String[] args){
String ss = "xarxatxm";
char first = ss.charAt(0);
char last == ss.charAt(3);
if(first == "x"){
String findx = ss.substring(0, 2);
}
if(last == "x"){
String findX = ss.substring(3, 5);
}
System.out.print(findx + findX);
}
}
Also, is there a way to implement the for loop to cycle through the String looking for x also?
I just need some advice to see where my code is going wrong.
You cannot find characters using charAt - it's for getting a character once you know where it is.
Is there a way to implement the for loop to cycle through the String looking for x also?
You need to use indexOf for finding positions of characters. Pass the initial position which is the position of the last x that you found so far to get the subsequent position.
For example, the code below
String s = "xarxatxm";
int pos = -1;
while (true) {
pos = s.indexOf('x', pos+1);
if (pos < 0) break;
System.out.println(pos);
}
prints 0 3 6 for the three positions of 'x' in the string.
Related
I'm new to java and I wrote this method to input a string word and output the word spelled backwards. The intent is to create a method and not use an already existing method such as the simple reverse. Please help point me in the direction of how to do this to reverse a word. I'm also trying to determine/count if there are palindromes. Please help! I've read other questions and I can't find anything specific enough to my case. I know that my code doesn't run, though I'm unsure how to fix it to get the correct output.
An example would be the word "backwards" to go to "sdrawkcab".
public static int reverseWord(String word) {
int palindromes = 0;
for (int i = word.length(); i >= 0; i--) {
System.out.print(i);
word.equalsIgnoreCase();
if (word.charAt(i)) == index(word.charAt(0 && 1))) {
palindromes++
System.out.println(palindromes)
}
return i;
}
}
There are multiple problems with your code.
1.The prototype of equalsIgnoreCase is
public boolean equalsIgnoreCase(String str);
So this method expect a String to be passed,but your not not passing anything here.To fix this,pass another string with whom you want to match your word like this..
word.equalsIgnoreCase("myAnotherString");
2.word.charAt(i);
Suppose word="qwerty",so indexing of each character will be like this
/* q w e r t y
0 1 2 3 4 5 */
So when you use i = word.length();i will 6 since word is of length 6.So
word.charAt(i) will search for character at index 6,but since there is not index 6,it will return an exception ArrayIndexOutOfBound.To fix this,start i from word.length()-1.
3.if (word.charAt(i));
This extra " ) ".Remove it.
Is Index() your own method?.If Yes,then check that also.
the below code prints the reverse of the input string and checks if it is a palindrome
public static void main(String[] args) {
String input = "dad";
char temp[] = input.toCharArray();//converting it to a array so that each character can be compared to the original string
char output[] = new char[temp.length];//taking another array of the same size as the input string
for (int i = temp.length - 1, j = 0; i >= 0; i--, j++) {//i variable for iterating through the input string and j variable for inserting data into output string.
System.out.print(temp[i]);//printing each variable of the input string in reverse order.
output[j] = temp[i];//inserting data into output string
}
System.out.println(String.valueOf(output));
if (String.valueOf(output).equalsIgnoreCase(input)) {//comparing the output string with the input string for palindrome check
System.out.println("palindrome");
}
}
Because your question about what is wrong with your code was already answered here is another way you could do it by using some concepts which are somewhat less low level than directly working with character arrays
public static boolean printWordAndCheckIfPalindrome(final String word) {
// Create a StringBuilder which helps when building a string
final StringBuilder reversedWordBuilder = new StringBuilder("");
// Get a stream of the character values of the word
word.chars()
// Add each character to the beginning of the reversed word,
// example for "backwards": "b", "ab", "cab", "kcab", ...
.forEach(characterOfString -> reversedWordBuilder.insert(0, (char) characterOfString));
// Generate a String out of the contents of the StringBuilder
final String reversedWord = reversedWordBuilder.toString();
// print the reversed word
System.out.println(reversedWord);
// if the reversed word equals the given word it is a palindrome
return word.equals(reversedWord);
}
Look for patterns like "zip" and "zap" in the string -- length-3, starting with 'z' and ending with 'p'. Return a string where for all such words, the middle letter is gone, so "zipXzap" yields "zpXzp"
Here is a solution i got from someone:
public class Rough {
public static void main(String [] args){
StringBuffer mat = new StringBuffer("matziplzdpaztp");
for(int i = 0; i < mat.length() - 2; ++i){
if (mat.charAt(i) == 'z' & mat.charAt(i + 2) == 'p'){
mat.deleteCharAt(i + 1);
}
}
System.out.println(mat);
}
}
But why is it that the for loop condition (i < mat.length() -2) is not (i < mat.length())????
Because in the loop:
if (mat.charAt(i) == 'z' & mat.charAt(i + 2) == 'p'){
// -----------------------------------^^^^^
If i were bound by i < mat.length(), then i + 2 would be out of bounds.
Because you don't have to reach the end of your sentence since your words are at least three letters long.
"2" stands for "the length except the first word",you just need to check all the positions in the string variable , and treat the positions as the first word of the substring , so just ignore the "length of the substring without the first word".
in your case , the length of "z*p" is 3, you just check all the position in the string , and treat the position as z to check something ,so just ignore "*p" ,which has length 2.
mat.length() will give length 14 and if you check for mat.charAt(i + 2) at the end it will give java.lang.StringIndexOutOfBoundsException because the string counts from index 0 not from 1. If you still want to use mat.length() you have to replace the AND '&' operator with short circuit AND '&&' operator in if condition.
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(5 answers)
Closed 8 years ago.
I'm trying to find every possible anagram of a string in Java - By this I mean that if I have a 4 character long word I want all the possible 3 character long words derived from it, all the 2 character long and all the 1 character long. The most straightforward way I tought of is to use two nested for loops and iterare over the string. This is my code as of now:
private ArrayList<String> subsets(String word){
ArrayList<String> s = new ArrayList<String>();
int length = word.length();
for (int c=0; c<length; c++){
for (int i=0; i<length-c; i++){
String sub = word.substring(c, c+i+1);
System.out.println(sub);
//if (!s.contains(sub) && sub!=null)
s.add(sub);
}
}
//java.util.Collections.sort(s, new MyComparator());
//System.out.println(s.toString());
return s;
}
My problem is that it works for 3 letter words, fun yelds this result (Don't mind the ordering, the word is processed so that I have a string with the letters in alphabetical order):
f
fn
fnu
n
nu
u
But when I try 4 letter words, it leaves something out, as in catq gives me:
a
ac
acq
acqt
c
cq
cqt
q
qt
t
i.e., I don't see the 3 character long word act - which is the one I'm looking for when testing this method. I can't understand what the problem is, and it's most likely a logical error I'm making when creating the substrings. If anyone can help me out, please don't give me the code for it but rather the reasoning behind your solution. This is a piece of coursework and I need to come up with the code on my own.
EDIT: to clear something out, for me acq, qca, caq, aqc, cqa, qac, etc. are the same thing - To make it even clearer, what happens is that the string gets sorted in alphabetical order, so all those permutations should come up as one unique result, acq. So, I don't need all the permutations of a string, but rather, given a 4 character long string, all the 3 character long ones that I can derive from it - that means taking out one character at a time and returning that string as a result, doing that for every character in the original string.
I hope I have made my problem a bit clearer
It's working fine, you just misspelled "caqt" as "acqt" in your tests/input.
(The issue is probably that you're sorting your input. If you want substrings, you have to leave the input unsorted.)
After your edits: see Generating all permutations of a given string Then just sort the individual letters, and put them in a set.
Ok, as you've already devised your own solution, I'll give you my take on it. Firstly, consider how big your result list is going to be. You're essentially taking each letter in turn, and either including it or not. 2 possibilities for each letter, gives you 2^n total results, where n is the number of letters. This of course includes the case where you don't use any letter, and end up with an empty string.
Next, if you enumerate every possibility with a 0 for 'include this letter' and a 1 for don't include it, taking your 'fnu' example you end up with:
000 - ''
001 - 'u'
010 - 'n'
011 - 'nu'
100 - 'f'
101 - 'fu' (no offense intended)
110 - 'fn'
111 - 'fnu'.
Clearly, these are just binary numbers, and you can derive a function that given any number from 0-7 and the three letter input, will calculate the corresponding subset.
It's fairly easy to do in java.. don't have a java compiler to hand, but this should be approximately correct:
public string getSubSet(string input, int index) {
// Should check that index >=0 and < 2^input.length here.
// Should also check that input.length <= 31.
string returnValue = "";
for (int i = 0; i < input.length; i++) {
if (i & (1 << i) != 0) // 1 << i is the equivalent of 2^i
returnValue += input[i];
}
return returnValue;
}
Then, if you need to you can just do a loop that calls this function, like this:
for (i = 1; i < (1 << input.length); i++)
getSubSet(input, i); // this doesn't do anything, but you can add it to a list, or output it as desired.
Note I started from 1 instead of 0- this is because the result at index 0 will be the empty string. Incidentally, this actually does the least significant bit first, so your output list would be 'f', 'n', 'fn', 'u', 'fu', 'nu', 'fnu', but the order didn't seem important.
This is the method I came up with, seems like it's working
private void subsets(String word, ArrayList<String> subset){
if(word.length() == 1){
subset.add(word);
return;
}
else {
String firstChar = word.substring(0,1);
word = word.substring(1);
subsets(word, subset);
int size = subset.size();
for (int i = 0; i < size; i++){
String temp = firstChar + subset.get(i);
subset.add(temp);
}
subset.add(firstChar);
return;
}
}
What I do is check if the word is bigger than one character, otherwise I'll add the character alone to the ArrayList and start the recursive process. If it is bigger, I save the first character and make a recursive call with the rest of the String. What happens is that the whole string gets sliced in characters saved in the recursive stack, until I hit the point where my word has become of length 1, only one character remaining.
When that happens, as I said at the start, the character gets added to the List, now the recursion starts and it looks at the size of the array, in the first iteration is 1, and then with a for loop adds the character saved in the stack for the previous call concatenated with every element in the ArrayList. Then it adds the character on its own and unwinds the recursion again.
I.E., with the word funthis happens:
f saved
List empty
recursive call(un)
-
u saved
List empty
recursive call(n)
-
n.length == 1
List = [n]
return
-
list.size=1
temp = u + list[0]
List = [n, un]
add the character saved in the stack on its own
List = [n, un, u]
return
-
list.size=3
temp = f + list[0]
List = [n, un, u, fn]
temp = f + list[1]
List = [n, un, u, fn, fun]
temp = f + list[2]
List = [n, un, u, fn, fun, fu]
add the character saved in the stack on its own
List = [n, un, u, fn, fun, fu, f]
return
I have been as clear as possible, I hope this clarifies what was my initial problem and how to solve it.
This is working code:
public static void main(String[] args) {
String input = "abcde";
Set<String> returnList = permutations(input);
System.out.println(returnList);
}
private static Set<String> permutations(String input) {
if (input.length() == 1) {
Set<String> a = new TreeSet<>();
a.add(input);
return a;
}
Set<String> returnSet = new TreeSet<>();
for (int i = 0; i < input.length(); i++) {
String prefix = input.substring(i, i + 1);
Set<String> permutations = permutations(input.substring(i + 1));
returnSet.add(prefix);
returnSet.addAll(permutations);
Iterator<String> it = permutations.iterator();
while (it.hasNext()) {
returnSet.add(prefix + it.next());
}
}
return returnSet;
}
public class newString {
public static void main (String args[]){
String title = "Book";
String title1;
title1 = title;
for(int i = 0; i < title.length(); i++){
for (int x = 0; x<title1.length(); x++){
if (title.charAt(i+x) == title1.charAt(x)){
System.out.print(title.charAt(0,1));
}
}
}
}
}
I really don't understand what I'm doing wrong here. What I need to do is define a string called "title", with "Book" in it, which I did, and create a second string called "title1". I need to create code to store the contents of title, into title1, but only every other character. For example: title1 should have "Bo" in it. What am I doing wrong?
Here's the looping solution with fewer operations. Instead of checking if i is even, just increment by 2.
String title1 = "Some title";
String title2 = "";
for (int i = 0; i < title1.length(); i += 2)
{
title2 += title1.charAt(i);
}
You algorithm is wrong, it seems what you need to do is to extract out every nth character from source string, for example:
String source = "Book";
End result should be "Bo"
The algorithm should be:
Iterate through each character in the original string, use a stride as you need, in this case, a stride of 2 should do (so rather than increment by one, increment by the required stride)
Take the character at that index and add it to your second string
The end result should be a string which holds every nth character.
I don't really understand what you are attempting, but I can tell you what you are doing. Your loop structure does the following:
when i = 0, it compares all characters in both strings (0 + n = n, so the inner loop goes from x - title1.length()).
when i = 1, compare all characters except the first one (for size x, 1 + n = x - 1 comparisons).
when i =2, compare x / 2 characters (for size x, 2 + n = x / 2)
when i = 3, compare x / 3 characters
... and so on
System.out.print(title.charAt(0,1)) Shouldn't even compile. charAt(int) is the correct call. And if title length is greater than 0, this will always print a String with a single character -- the first one in title. And it will always be the same unless you reassign title to a different String.
Also this code will always throw an IndexOutOfBoundsException at title.charAt(i+x) when i = title.length() - 1 and x > 0.
This question already has an answer here:
Closed 10 years ago.
Possible Duplicate:
Programming java to determine a symmetrical word
am new here, but I am having hard time figuring out how to write a code to determine an input of word and see if the first is matching with the end of the word. You may input abba and get answer it's evenly symmetric and aba is oddly symmetric.
Please show me how:(
Just two main things.
first I want to know if it's oddly or evenly amount of letter(number of letter divided by 2,if it's ending with 0.5, it's oddly symmetric, if is an integer it's evenly symmetric.
second I want to get (i.e 1=n,2=n-1,3=n-2...) position of the letter in the word to be the main idea of the execution.If there is a last letter in the oddly symmetric word, ignore the last remaining letter.
I appreciate any headstart or idea:) Thanks!
Thanks KDiTraglia, I made the code and compiled and here is what I put. I am not getting any further.
Reported problem:
Exception in thread "main" java.lang.Error: Unresolved compilation problems: reverse cannot be resolved or is not a field reverse cannot be resolved or is not a field Syntax error, insert ") Statement" to complete IfStatement
This is what i got from, KDiTraglia's help
public class WordSymmetric {
public static void main(String[] args) {
String word = "abccdccba";
if ( (word.length() % 2) == 1 ) {
System.out.println("They are oddly symmetric");
//odd
}
else {
System.out.println("They are evenly symmetric");
//even
}
int halfLength = word.length() / 2;
String firstHalf = word.substring(0, halfLength);
String secondHalf = word.substring(halfLength, word.length());
System.out.println(secondHalf.reverse());
if (firstHalf.equals(secondHalf.reverse()) {
System.out.println("They match");
//they match
}
} }
String does not have a reverse method. You could use the apache commons lang library for this purpose:
http://commons.apache.org/lang/api-release/org/apache/commons/lang3/StringUtils.html#reverse%28java.lang.String%29
The reverse() approach is very clean and readable. Unfortunately there is no reverse() method for Strings. So you would either have to take an external library (StringUtils from the appache common lang3 library has a reverse method) or code it yourself.
public static String reverse(String inputString) {
StringBuilder reverseString = new StringBuilder();
for(int i = inputString.length(); i > 0; --i) {
char result = inputString.charAt(i-1);
reverseString.append(result);
}
return reverseString.toString();
}
(This only works for characters that can fit into a char. So if you need something more general, you would have to expand it.)
Then you can just have a method like this:
enum ePalindromResult { NO_PALINDROM, PALINDROM_ODD, PALINDROM_EVEN };
public static ePalindromResult checkForPalindrom(String inputStr) {
// this uses the org.apache.commons.lang3.StringUtils class:
if (inputStr.equals(StringUtils.reverse(inputStr)) {
if (inputStr.length % 2 == 0) return PALINDROM_EVEN;
else return PALINDROM_ODD;
} else return NO_PALINDROM;
}
System.out.println(secondHalf.reverse());
There is no reverse() method defined fro String
I would probably loop over word from index 0 to the half (word.length() / 2) and compare the character at the current index (word.charAt(i)) with the correspoding from the other half (word.charAt(word.length() - i).
This is just a rough draft, you probably need to think about the loop end index, depending on oddly or evenly symmetry.
You can adapt this :
final char[] word = "abccdccba".toCharArray(); // work also with "abccccba"
final int t = word.length;
boolean ok = true;
for (int i = t / 2; i > 0; i--) {
if (word[i - 1] != word[t - i]) {
ok = false;
break;
}
System.out.println(word[i - 1] + "\t" + word[word.length - i]);
}
System.out.println(ok);
Console :
c c
c c
b b
a a
true
Use class StringBuffer instead of String