I am trying to create a method which will either remove all duplicates from a string or only keep the same 2 characters in a row based on a parameter.
For example:
helllllllo -> helo
or
helllllllo -> hello - This keeps double letters
Currently I remove duplicates by doing:
private String removeDuplicates(String word) {
StringBuffer buffer = new StringBuffer();
for (int i = 0; i < word.length(); i++) {
char letter = word.charAt(i);
if (buffer.length() == 0 && letter != buffer.charAt(buffer.length() - 1)) {
buffer.append(letter);
}
}
return buffer.toString();
}
If I want to keep double letters I was thinking of having a method like private String removeDuplicates(String word, boolean doubleLetter)
When doubleLetter is true it will return hello not helo
I'm not sure of the most efficient way to do this without duplicating a lot of code.
why not just use a regex?
public class RemoveDuplicates {
public static void main(String[] args) {
System.out.println(new RemoveDuplicates().result("hellllo", false)); //helo
System.out.println(new RemoveDuplicates().result("hellllo", true)); //hello
}
public String result(String input, boolean doubleLetter){
String pattern = null;
if(doubleLetter) pattern = "(.)(?=\\1{2})";
else pattern = "(.)(?=\\1)";
return input.replaceAll(pattern, "");
}
}
(.) --> matches any character and puts in group 1.
?= --> this is called a positive lookahead.
?=\\1 --> positive lookahead for the first group
So overall, this regex looks for any character that is followed (positive lookahead) by itself. For example aa or bb, etc. It is important to note that only the first character is part of the match actually, so in the word 'hello', only the first l is matched (the part (?=\1) is NOT PART of the match). So the first l is replaced by an empty String and we are left with helo, which does not match the regex
The second pattern is the same thing, but this time we look ahead for TWO occurrences of the first group, for example helllo. On the other hand 'hello' will not be matched.
Look here for a lot more: Regex
P.S. Fill free to accept the answer if it helped.
try
String s = "helllllllo";
System.out.println(s.replaceAll("(\\w)\\1+", "$1"));
output
helo
Taking this previous SO example as a starting point, I came up with this:
String str1= "Heelllllllllllooooooooooo";
String removedRepeated = str1.replaceAll("(\\w)\\1+", "$1");
System.out.println(removedRepeated);
String keepDouble = str1.replaceAll("(\\w)\\1{2,}", "$1");
System.out.println(keepDouble);
It yields:
Helo
Heelo
What it does:
(\\w)\\1+ will match any letter and place it in a regex capture group. This group is later accessed through the \\1+. Meaning that it will match one or more repetitions of the previous letter.
(\\w)\\1{2,} is the same as above the only difference being that it looks after only characters which are repeated more than 2 times. This leaves the double characters untouched.
EDIT:
Re-read the question and it seems that you want to replace multiple characters by doubles. To do that, simply use this line:
String keepDouble = str1.replaceAll("(\\w)\\1+", "$1$1");
Try this, this will be most efficient way[Edited after comment]:
public static String removeDuplicates(String str) {
int checker = 0;
StringBuffer buffer = new StringBuffer();
for (int i = 0; i < str.length(); ++i) {
int val = str.charAt(i) - 'a';
if ((checker & (1 << val)) == 0)
buffer.append(str.charAt(i));
checker |= (1 << val);
}
return buffer.toString();
}
I am using bits to identify uniqueness.
EDIT:
Whole logic is that if a character has been parsed then its corrresponding bit is set and next time when that character comes up then it will not be added in String Buffer the corresponding bit is already set.
Related
I have the following problem which states
Replace all characters in a string with + symbol except instances of the given string in the method
so for example if the string given was abc123efg and they want me to replace every character except every instance of 123 then it would become +++123+++.
I figured a regular expression is probably the best for this and I came up with this.
str.replaceAll("[^str]","+")
where str is a variable, but its not letting me use the method without putting it in quotations. If I just want to replace the variable string str how can I do that? I ran it with the string manually typed and it worked on the method, but can I just input a variable?
as of right now I believe its looking for the string "str" and not the variable string.
Here is the output its right for so many cases except for two :(
List of open test cases:
plusOut("12xy34", "xy") → "++xy++"
plusOut("12xy34", "1") → "1+++++"
plusOut("12xy34xyabcxy", "xy") → "++xy++xy+++xy"
plusOut("abXYabcXYZ", "ab") → "ab++ab++++"
plusOut("abXYabcXYZ", "abc") → "++++abc+++"
plusOut("abXYabcXYZ", "XY") → "++XY+++XY+"
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
plusOut("--++ab", "++") → "++++++"
plusOut("aaxxxxbb", "xx") → "++xxxx++"
plusOut("123123", "3") → "++3++3"
Looks like this is the plusOut problem on CodingBat.
I had 3 solutions to this problem, and wrote a new streaming solution just for fun.
Solution 1: Loop and check
Create a StringBuilder out of the input string, and check for the word at every position. Replace the character if doesn't match, and skip the length of the word if found.
public String plusOut(String str, String word) {
StringBuilder out = new StringBuilder(str);
for (int i = 0; i < out.length(); ) {
if (!str.startsWith(word, i))
out.setCharAt(i++, '+');
else
i += word.length();
}
return out.toString();
}
This is probably the expected answer for a beginner programmer, though there is an assumption that the string doesn't contain any astral plane character, which would be represented by 2 char instead of 1.
Solution 2: Replace the word with a marker, replace the rest, then restore the word
public String plusOut(String str, String word) {
return str.replaceAll(java.util.regex.Pattern.quote(word), "#").replaceAll("[^#]", "+").replaceAll("#", word);
}
Not a proper solution since it assumes that a certain character or sequence of character doesn't appear in the string.
Note the use of Pattern.quote to prevent the word being interpreted as regex syntax by replaceAll method.
Solution 3: Regex with \G
public String plusOut(String str, String word) {
word = java.util.regex.Pattern.quote(word);
return str.replaceAll("\\G((?:" + word + ")*+).", "$1+");
}
Construct regex \G((?:word)*+)., which does more or less what solution 1 is doing:
\G makes sure the match starts from where the previous match leaves off
((?:word)*+) picks out 0 or more instance of word - if any, so that we can keep them in the replacement with $1. The key here is the possessive quantifier *+, which forces the regex to keep any instance of the word it finds. Otherwise, the regex will not work correctly when the word appear at the end of the string, as the regex backtracks to match .
. will not be part of any word, since the previous part already picks out all consecutive appearances of word and disallow backtrack. We will replace this with +
Solution 4: Streaming
public String plusOut(String str, String word) {
return String.join(word,
Arrays.stream(str.split(java.util.regex.Pattern.quote(word), -1))
.map((String s) -> s.replaceAll("(?s:.)", "+"))
.collect(Collectors.toList()));
}
The idea is to split the string by word, do the replacement on the rest, and join them back with word using String.join method.
Same as above, we need Pattern.quote to avoid split interpreting the word as regex. Since split by default removes empty string at the end of the array, we need to use -1 in the second parameter to make split leave those empty strings alone.
Then we create a stream out of the array and replace the rest as strings of +. In Java 11, we can use s -> String.repeat(s.length()) instead.
The rest is just converting the Stream to an Iterable (List in this case) and joining them for the result
This is a bit trickier than you might initially think because you don't just need to match characters, but the absence of specific phrase - a negated character set is not enough. If the string is 123, you would need:
(?<=^|123)(?!123).*?(?=123|$)
https://regex101.com/r/EZWMqM/1/
That is - lookbehind for the start of the string or "123", make sure the current position is not followed by 123, then lazy-repeat any character until lookahead matches "123" or the end of the string. This will match all characters which are not in a "123" substring. Then, you need to replace each character with a +, after which you can use appendReplacement and a StringBuffer to create the result string:
String inputPhrase = "123";
String inputStr = "abc123efg123123hij";
StringBuffer resultString = new StringBuffer();
Pattern regex = Pattern.compile("(?<=^|" + inputPhrase + ")(?!" + inputPhrase + ").*?(?=" + inputPhrase + "|$)");
Matcher m = regex.matcher(inputStr);
while (m.find()) {
String replacement = m.group(0).replaceAll(".", "+");
m.appendReplacement(resultString, replacement);
}
m.appendTail(resultString);
System.out.println(resultString.toString());
Output:
+++123+++123123+++
Note that if the inputPhrase can contain character with a special meaning in a regular expression, you'll have to escape them first before concatenating into the pattern.
You can do it in one line:
input = input.replaceAll("((?:" + str + ")+)?(?!" + str + ").((?:" + str + ")+)?", "$1+$2");
This optionally captures "123" either side of each character and puts them back (a blank if there's no "123"):
So instead of coming up with a regular expression that matches the absence of a string. We might as well just match the selected phrase and append + the number of skipped characters.
StringBuilder sb = new StringBuilder();
Matcher m = Pattern.compile(Pattern.quote(str)).matcher(input);
while (m.find()) {
for (int i = 0; i < m.start(); i++) sb.append('+');
sb.append(str);
}
int remaining = input.length() - sb.length();
for (int i = 0; i < remaining; i++) {
sb.append('+');
}
Absolutely just for the fun of it, a solution using CharBuffer (unexpectedly it took a lot more that I initially hoped for):
private static String plusOutCharBuffer(String input, String match) {
int size = match.length();
CharBuffer cb = CharBuffer.wrap(input.toCharArray());
CharBuffer word = CharBuffer.wrap(match);
int x = 0;
for (; cb.remaining() > 0;) {
if (!cb.subSequence(0, size < cb.remaining() ? size : cb.remaining()).equals(word)) {
cb.put(x, '+');
cb.clear().position(++x);
} else {
cb.clear().position(x = x + size);
}
}
return cb.clear().toString();
}
To make this work you need a beast of a pattern. Let's say you you are operating on the following test case as an example:
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
What you need to do is build a series of clauses in your pattern to match a single character at a time:
Any character that is NOT "X", "Y" or "Z" -- [^XYZ]
Any "X" not followed by "YZ" -- X(?!YZ)
Any "Y" not preceded by "X" -- (?<!X)Y
Any "Y" not followed by "Z" -- Y(?!Z)
Any "Z" not preceded by "XY" -- (?<!XY)Z
An example of this replacement can be found here: https://regex101.com/r/jK5wU3/4
Here is an example of how this might work (most certainly not optimized, but it works):
import java.util.regex.Pattern;
public class Test {
public static void plusOut(String text, String exclude) {
StringBuilder pattern = new StringBuilder("");
for (int i=0; i<exclude.length(); i++) {
Character target = exclude.charAt(i);
String prefix = (i > 0) ? exclude.substring(0, i) : "";
String postfix = (i < exclude.length() - 1) ? exclude.substring(i+1) : "";
// add the look-behind (?<!X)Y
if (!prefix.isEmpty()) {
pattern.append("(?<!").append(Pattern.quote(prefix)).append(")")
.append(Pattern.quote(target.toString())).append("|");
}
// add the look-ahead X(?!YZ)
if (!postfix.isEmpty()) {
pattern.append(Pattern.quote(target.toString()))
.append("(?!").append(Pattern.quote(postfix)).append(")|");
}
}
// add in the other character exclusion
pattern.append("[^" + Pattern.quote(exclude) + "]");
System.out.println(text.replaceAll(pattern.toString(), "+"));
}
public static void main(String [] args) {
plusOut("12xy34", "xy");
plusOut("12xy34", "1");
plusOut("12xy34xyabcxy", "xy");
plusOut("abXYabcXYZ", "ab");
plusOut("abXYabcXYZ", "abc");
plusOut("abXYabcXYZ", "XY");
plusOut("abXYxyzXYZ", "XYZ");
plusOut("--++ab", "++");
plusOut("aaxxxxbb", "xx");
plusOut("123123", "3");
}
}
UPDATE: Even this doesn't quite work because it can't deal with exclusions that are just repeated characters, like "xx". Regular expressions are most definitely not the right tool for this, but I thought it might be possible. After poking around, I'm not so sure a pattern even exists that might make this work.
The problem in your solution that you put a set of instance string str.replaceAll("[^str]","+") which it will exclude any character from the variable str and that will not solve your problem
EX: when you try str.replaceAll("[^XYZ]","+") it will exclude any combination of character X , character Y and character Z from your replacing method so you will get "++XY+++XYZ".
Actually you should exclude a sequence of characters instead in str.replaceAll.
You can do it by using capture group of characters like (XYZ) then use a negative lookahead to match a string which does not contain characters sequence : ^((?!XYZ).)*$
Check this solution for more info about this problem but you should know that it may be complicated to find regular expression to do that directly.
I have found two simple solutions for this problem :
Solution 1:
You can implement a method to replace all characters with '+' except the instance of given string:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
for(int i = 0; i < str.length(); i++){
// exclude any instance string of exWord from replacing process in str
if(str.substring(i, str.length()).indexOf(exWord) + i == i){
i = i + exWord.length()-1;
}
else{
str = str.substring(0,i) + "+" + str.substring(i+1);//replace each character with '+' symbol
}
}
Note : str.substring(i, str.length()).indexOf(exWord) + i this if statement will exclude any instance string of exWord from replacing process in str.
Output:
+++++++XYZ
Solution 2:
You can try this Approach using ReplaceAll method and it doesn't need any complex regular expression:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
str = str.replaceAll(exWord,"*"); // replace instance string with * symbol
str = str.replaceAll("[^*]","+"); // replace all characters with + symbol except *
str = str.replaceAll("\\*",exWord); // replace * symbol with instance string
Note : This solution will work only if your input string str doesn't contain any * symbol.
Also you should escape any character with a special meaning in a regular expression in phrase instance string exWord like : exWord = "++".
Method Name - getCharsThatFollowPattern
Prompt and what its supposed to do. Takes a String text and a String pattern as parameters, and returns an ArrayList. The returned list should contain the character that follows each non-tail occurrence of the pattern in the text. (A non-tail occurrence of the pattern is one that is not at the very end of the text.) The length of the list must be the same as the number of non-tail occurrences of the pattern. The character stored at index n of the list must be the character that followed the nth non-tail occurrence of the pattern. For example, getCharsThatFollowPattern("abcabdabcab", "ab") should return the ArrayList ['c', 'd', 'c'].
My code is below along with the results.
public static ArrayList<Character> getCharsThatFollowPattern (String text, String pattern)
{
ArrayList<Character> character = new ArrayList<>();
// String str = text;
// String findStr = pattern;
int lastIndex = 0;
while (lastIndex != -1) {
lastIndex = text.indexOf(pattern, lastIndex);
if (lastIndex != -1) {
lastIndex += pattern.length();
char c = text.charAt(text.lastIndexOf(pattern) - 1);
character.add(c);
}
}
return character;
}
The results of the code are as follow.
getCharsThatFollowPattern("abcabdabcab", "ab") returns [c,c,c,c];
- which is wrong.
getCharsThatFollowPattern("abababa", "aba") returns [b, b];
- which is what its supposed to do.
My teacher supplied me with a test case that works but when I tested it myself it does not work.
#Test
public void testGetCharsThatFollowPattern ()
{
ArrayList<Character> list = new ArrayList<Character>();
list.add('b');
list.add('b');
assertEquals(list,
PS5Library.getCharsThatFollowPattern("abababa", "aba"));
}
}
I have been scavenging stack overflow and cannot find an answer to my question. Any insights on to why how to fix this would be great. Thank you guys.
I think you just have a small math error and just need to change the line that assigns c to this:
char c = text.charAt(lastIndex);
You already accepted an answer but the reason why you always added c is the following line:
char c = text.charAt(text.lastIndexOf(pattern) - 1);
You always look for the character after the last occurrence of the pattern. The StringIndexOutOfBoundsException results from the fact that the text ends with the pattern without subsequent character. The index itself is zero-indexed, so charAt(5) tries to access the 6th character of a five-character-long text.
You might ignore the exception or check the length of the text with the index and only call charAt(index) if index < text.length(). The latter would prevent unnecessary creations of Exceptions and it's better style, too ;-)
There's a string
String str = "ggg;ggg;nnn;nnn;aaa;aaa;xxx;xxx;";
How do I split it into strings like this
"ggg;ggg;"
"nnn;nnn;"
"aaa;aaa;"
"xxx;xxx;"
???????
Using Regex
String input = "ggg;ggg;nnn;nnn;aaa;aaa;xxx;xxx;";
Pattern p = Pattern.compile("([a-z]{3});\\1;");
Matcher m = p.matcher(input);
while (m.find())
// m.group(0) is the result
System.out.println(m.group(0));
Will output
ggg;ggg;
nnn;nnn;
aaa;aaa;
xxx;xxx;
I assume that the you only want to check if the last segment is similar and not every segment that has been read.
If that is not the case then you would probably have to use an ArrayList instead of a Stack.
I also assumed that each segment has the format /([a-z])\1\1/.
If that is not the case either then you should change the if statement with:
(stack.peek().substring(0,index).equals(temp))
public static Stack<String> splitString(String text, char split) {
Stack<String> stack = new Stack<String>();
int index = text.indexOf(split);
while (index != -1) {
String temp = text.substring(0, index);
if (!stack.isEmpty()) {
if (stack.peek().charAt(0) == temp.charAt(0)) {
temp = stack.pop() + split + temp;
}
}
stack.push(temp);
text = text.substring(index + 1);
index = text.indexOf(split);
}
return stack;
}
Split and join them.
public static void main(String[] args) throws Exception {
String data = "ggg;ggg;nnn;nnn;aaa;aaa;xxx;xxx;";
String del = ";";
int splitSize = 2;
StringBuilder sb = new StringBuilder();
for (Iterable<String> iterable : Iterables.partition(Splitter.on(del).split(data), splitSize)) {
sb.append("\"").append(Joiner.on(del).join(iterable)).append(";\"");
}
sb.delete(sb.length()-3, sb.length());
System.out.println(sb.toString());
}
Ref : Split a String at every 3rd comma in Java
Use split with a regex:
String data="ggg;ggg;nnn;nnn;aaa;aaa;xxx;xxx;";
String [] array=data.split("(?<=\\G\\S\\S\\S;\\S\\S\\S);");
S: A non-whitespace character
G: last match/start of string, think of it of a way to skip delimiting if the
previous string matches current one.
?<=:positive look-behind will match semicolon which has string behind it.
Some other answer, that only works given your specific example input.
You see, in your example, there are two similarities:
All patterns seem to have exactly three characters
All patterns occur exactly twice
In other words: if those two properties are really met for all your input, you could avoid splitting - as you know exactly what to find in each position of your string.
Of course, following the other answers for "real" splitting are more flexible; but (theoretically), you could just go forward and do a bunch of substring calls in order to directly access all elements.
I have inputs like
AS23456SDE
MFD324FR
I need to get First Character values like
AS, MFD
There should no first two or first 3 characters input can be changed. Need to get first characters before a number.
Thank you.
Edit : This is what I have tried.
public static String getPrefix(String serial) {
StringBuilder prefix = new StringBuilder();
for(char c : serial.toCharArray()){
if(Character.isDigit(c)){
break;
}
else{
prefix.append(c);
}
}
return prefix.toString();
}
Here is a nice one line solution. It uses a regex to match the first non numeric characters in the string, and then replaces the input string with this match.
public String getFirstLetters(String input) {
return new String("A" + input).replaceAll("^([^\\d]+)(.*)$", "$1")
.substring(1);
}
System.out.println(getFirstLetters("AS23456SDE"));
System.out.println(getFirstLetters("1AS123"));
Output:
AS
(empty)
A simple solution could be like this:
public static void main (String[]args) {
String str = "MFD324FR";
char[] characters = str.toCharArray();
for(char c : characters){
if(Character.isDigit(c))
break;
else
System.out.print(c);
}
}
Use the following function to get required output
public String getFirstChars(String str){
int zeroAscii = '0'; int nineAscii = '9';
String result = "";
for (int i=0; i< str.lenght(); i++){
int ascii = str.toCharArray()[i];
if(ascii >= zeroAscii && ascii <= nineAscii){
result = result + str.toCharArray()[i];
}else{
return result;
}
}
return str;
}
pass your string as argument
I think this can be done by a simple regex which matches digits and java's string split function. This Regex based approach will be more efficient than the methods using more complicated regexs.
Something as below will work
String inp = "ABC345.";
String beginningChars = inp.split("[\\d]+",2)[0];
System.out.println(beginningChars); // only if you want to print.
The regex I used "[\\d]+" is escaped for java already.
What it does?
It matches one or more digits (d). d matches digits of any language in unicode, (so it matches japanese and arabian numbers as well)
What does String beginningChars = inp.split("[\\d]+",2)[0] do?
It applies this regex and separates the string into string arrays where ever a match is found. The [0] at the end selects the first result from that array, since you wanted the starting chars.
What is the second parameter to .split(regex,int) which I supplied as 2?
This is the Limit parameter. This means that the regex will be applied on the string till 1 match is found. Once 1 match is found the string is not processed anymore.
From the Strings javadoc page:
The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array. If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter. If n is non-positive then the pattern will be applied as many times as possible and the array can have any length. If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.
This will be efficient if your string is huge.
Possible other regex if you want to split only on english numerals
"[0-9]+"
public static void main(String[] args) {
String testString = "MFD324FR";
int index = 0;
for (Character i : testString.toCharArray()) {
if (Character.isDigit(i))
break;
index++;
}
System.out.println(testString.substring(0, index));
}
this prints the first 'n' characters before it encounters a digit (i.e. integer).
I have some raw output that I want to clean up and make presentable but right now I go about it in a very ugly and cumbersome way, I wonder if anyone might know a clean and elegant way in which to perform the same operation.
int size = charOutput.size();
for (int i = size - 1; i >= 1; i--)
{
if(charOutput.get(i).compareTo(charOutput.get(i - 1)) == 0)
{
charOutput.remove(i);
}
}
for(int x = 0; x < charOutput.size(); x++)
{
if(charOutput.get(x) == '?')
{
charOutput.remove(x);
}
}
String firstOne = Arrays.toString(charOutput.toArray());
String secondOne = firstOne.replaceAll(",","");
String thirdOne = secondOne.substring(1, secondOne.length() - 1);
String output = thirdOne.replaceAll(" ","");
return output;
ZouZou has the right code for fixing the final few calls in your code. I have some suggestions for the for loops. I hope I got them right...
These work after you get the String represented by charOutput, using a method such as the one suggested by ZouZou.
Your first block appears to remove all repeated letters. You can use a regular expression for that:
Pattern removeRepeats = Pattern.compile("(.)\\1{1,}");
// "(.)" creates a group that matches any character and puts it into a group
// "\\1" gets converted to "\1" which is a reference to the first group, i.e. the character that "(.)" matched
// "{1,}" means "one or more"
// So the overall effect is "one or more of a single character"
To use:
removeRepeats.matcher(s).replaceAll("$1");
// This creates a Matcher that matches the regex represented by removeRepeats to the contents of s, and replaces the parts of s that match the regex represented by removeRepeats with "$1", which is a reference to the first group captured (i.e. "(.)", which is the first character matched"
To remove the question mark, just do
Pattern removeQuestionMarks = Pattern.compile("\\?");
// Because "?" is a special symbol in regex, you have to escape it with a backslash
// But since backslashes are also a special symbol, you have to escape the backslash too.
And then to use, do the same thing as was done above except with replaceAll("");
And you're done!
If you really wanted to, you can combine a lot of regex into two super-regex expressions (and one normal regex expression):
Pattern p0 = Pattern.compile("(\\[|\\]|\\,| )"); // removes brackets, commas, and spaces
Pattern p1 = Pattern.compile("(.)\\1{1,}"); // Removes duplicate characters
Pattern p2 = Pattern.compile("\\?");
String removeArrayCharacters = p0.matcher(charOutput.toString()).replaceAll("");
String removeDuplicates = p1.matcher(removeArrayCharacters).replaceAll("$1");
return p2.matcher(removeDuplicates).replaceAll("");
Use a StringBuilder and append each character you want, at the end just return myBuilder.toString();
Instead of this:
String firstOne = Arrays.toString(charOutput.toArray());
String secondOne = firstOne.replaceAll(",","");
String thirdOne = secondOne.substring(1, secondOne.length() - 1);
String output = thirdOne.replaceAll(" ","");
return output;
Simply do:
StringBuilder sb = new StringBuilder();
for(Character c : charOutput){
sb.append(c);
}
return sb.toString();
Note that you are doing a lot of unnecessary work (by iterating through the list and removing some elements). What you can actually do is just iterate one time and then if the condition fullfits your requirements (the two adjacent characters are not the same and no question mark) then append it to the StringBuilder directly.
This task could also be a job for a regular expression.
If you don't want to use Regex try this version to remove consecutive characters and '?':
int size = charOutput.size();
if (size == 1) return Character.toString((Character)charOutput.get(0));
else if (size == 0) return null;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < size - 1; i++) {
Character temp = (Character)charOutput.get(i);
if (!temp.equals(charOutput.get(i+1)) && !temp.equals('?'))
sb.append(temp);
}
//for the last element
if (!charOutput.get(size-1).equals(charOutput.get(size-2))
&& !charOutput.get(size-1).equals('?'))
sb.append(charOutput.get(size-1));
return sb.toString();