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Strip Leading and Trailing Spaces From Java String
(8 answers)
Closed 4 years ago.
I am writing this function for a J2ME application, so I don't have some of the more advanced / modern Java classes available to me. I am getting java.lang.ArrayIndexOutOfBoundsException on this. So, apparently either it doesn't like the way I've initialized the newChars array, or I'm not doing something correctly when calling System.arraycopy.
/*
* remove any leading and trailing spaces
*/
public static String trim(String str) {
char[] chars = str.toCharArray();
int len = chars.length;
// leading
while ( (len > 0 ) && ( chars[0] == ' ' ) ) {
char[] newChars = new char[] {}; // initialize empty array
System.arraycopy(chars, 1, newChars, 0, len - 1);
chars = newChars;
len = chars.length;
}
// TODO: trailing
return chars.toString();
}
The simple way to trim leading and trailing whitespace is to call String.trim(). With Java 11 and later, you can also use String.strip() which uses a different interpretation of what "white space" means1.
If you just want to trim just leading and trailing spaces (rather than all leading and trailing whitespace), there is an Apache commons method called StringUtils.strip(String, String) that can do this; call it with " " as the 2nd argument.
Your attempted code has a number of bugs, and is fundamentally inefficient. If you really want to implement this yourself, then you should:
count the leading space characters
count the trailing space characters
if either count is non-zero, call String.substring(from, end) to create a new string containing the characters you want to keep.
This approach avoids copying any characters2.
1 - The different meanings are explained in the respective javadocs. Alternatively, read the answers to Difference between String trim() and strip() methods in Java 11.
2 - Actually, that depends on the implementation of String. For some implementations there will be no copying, for others a single copy is made. But either is an improvement on your approach, which entails a minimum of 2 copies, and more if there are any characters to trim.
String.trim() is very old, at least to java 1.3. You don't have this?
Apache StringUtils.strip is the best answer here that works with all expected white space characters (not just space), and can be downloaded here:
Here's the relevant code ripped from this source file to implement it in your own class if you wanted, but really, just download and use StringUtils to get more bang for your buck! Note that you can use StringUtils.stripStart to trim any leading character from a java string as well.
public static final int INDEX_NOT_FOUND = -1
public static String strip(final String str) {
return strip(str, null);
}
public static String stripStart(final String str, final String stripChars) {
int strLen;
if (str == null || (strLen = str.length()) == 0) {
return str;
}
int start = 0;
if (stripChars == null) {
while (start != strLen && Character.isWhitespace(str.charAt(start))) {
start++;
}
} else if (stripChars.isEmpty()) {
return str;
} else {
while (start != strLen && stripChars.indexOf(str.charAt(start)) != INDEX_NOT_FOUND) {
start++;
}
}
return str.substring(start);
}
public static String stripEnd(final String str, final String stripChars) {
int end;
if (str == null || (end = str.length()) == 0) {
return str;
}
if (stripChars == null) {
while (end != 0 && Character.isWhitespace(str.charAt(end - 1))) {
end--;
}
} else if (stripChars.isEmpty()) {
return str;
} else {
while (end != 0 && stripChars.indexOf(str.charAt(end - 1)) != INDEX_NOT_FOUND) {
end--;
}
}
return str.substring(0, end);
}
public static String strip(String str, final String stripChars) {
if (str == null || str.length() == 0) {
return str;
}
str = stripStart(str, stripChars);
return stripEnd(str, stripChars);
}
First of all, what others said about String.trim(). Really, don't reinvent the wheel.
But for the record, what's going wrong with your code is that Java arrays aren't resizeable. When you initially set up your target array, you create it as a size 0 array. You then tell System.arraycopy to stuff len - 1 characters in there. That's not going to work. If you wanted it to work, you'd need to set up the array as:
char[] newChars = new char[len - 1];
But that's amazingly inefficient, reallocating and copying a new array each time through the loop. Use the three steps that Stephen C mentioned, ending with a substring.
With JDK/11, now you can make use of the String.strip API to return a string whose value is this string, with all leading and trailing whitespace removed. The javadoc for the same is :
/**
* Returns a string whose value is this string, with all leading
* and trailing {#link Character#isWhitespace(int) white space}
* removed.
* <p>
* If this {#code String} object represents an empty string,
* or if all code points in this string are
* {#link Character#isWhitespace(int) white space}, then an empty string
* is returned.
* <p>
* Otherwise, returns a substring of this string beginning with the first
* code point that is not a {#link Character#isWhitespace(int) white space}
* up to and including the last code point that is not a
* {#link Character#isWhitespace(int) white space}.
* <p>
* This method may be used to strip
* {#link Character#isWhitespace(int) white space} from
* the beginning and end of a string.
*
* #return a string whose value is this string, with all leading
* and trailing white space removed
*
* #see Character#isWhitespace(int)
*
* #since 11
*/
public String strip()
The sample cases for these could be:--
System.out.println("".strip());
System.out.println(" both ".strip());
System.out.println(" leading".strip());
System.out.println("trailing ".strip());
If you don't want to use String.trim() method, then it can be implemented like below. The logic will handle different scenarios like space, tab and other special characters.
public static String trim(String str){
int i=0;
int j = str.length();
char[] charArray = str.toCharArray();
while((i<j) && charArray[i] <=' '){
i++;
}
while((i<j) && charArray[j-1]<= ' '){
j--;
}
return str.substring(i, j+1);
}
public static void main(String[] args) {
System.out.println(trim(" abcd ght trip "));
}
The destination array newChars is not large enough to hold the values copied. You need to initialize it to the length of the data you intend to copy (so, length - 1).
You can use Guava CharMatcher.
String outputString = CharMatcher.whitespace().trimFrom(inputString);
Note: This works because whitespace is all in the BMP.
Related
String.trim() in java removes all characters whose ascii value is less than or equal to 20 (space).
Any idea why Java did that instead of removing only space (ascii char 20)
public String trim() {
int len = count;
int st = 0;
int off = offset; /* avoid getfield opcode */
char[] val = value; /* avoid getfield opcode */
while ((st < len) && (val[off + st] <= ' ')) {
st++;
}
while ((st < len) && (val[off + len - 1] <= ' ')) {
len--;
}
return ((st > 0) || (len < count)) ? substring(st, len) : this;
}
Because there are many different ways of having empty space, besides that " " space character. Quoting the javadoc:
Returns a copy of the string, with leading and trailing whitespace omitted.
The javadoc is clear here: it is not about space but white space. Things that would show up as "empty" - but that are in fact different from plain " " empty strings.
In other words: this is a convenience method. Such methods are designed to provide that functionality that users do need/expect.
It would be absolutely counter-intuition to provide a trim() method that only works spaces.
A very typical scenario is: you receive some string. It could be entered by a user, it could be read from a file and represent a whole line. You are not interested in any trailing tabs, spaces, new line characters. Thus the fathers of the Java language give you a method to get rid of all these different characters easily. Instead of you calling trimSpaces(), trimTabs(), trimNewLines(), etc. pp.
The ASCII character code for space is actually 32, not 20. But if you look at the sorts of characters which come before 32 you will find many types of whitespace, such as tab and carriage return. The asssumption is the average user would want to strip all such whitespace surrounding a string.
To round out the answer given by #GhostCat here is a one-liner you can use to selectively trim only space:
String input = " Hello World! ";
input = input.replaceAll("[ ]*(.*)[ ]*", "$1");
The below one-liner works. The one given by #Tim Biegeleisen doesn't remove a trailing space.
String input = " Hello World! ";
input = input.replaceFirst("^\\s++", "").replaceFirst("\\s++$","");
I want to remove the leading and trailing whitespace from string:
String s = " Hello World ";
I want the result to be like:
s == "Hello world";
s.trim()
see String#trim()
Without any internal method, use regex like
s.replaceAll("^\\s+", "").replaceAll("\\s+$", "")
or
s.replaceAll("^\\s+|\\s+$", "")
or just use pattern in pure form
String s=" Hello World ";
Pattern trimmer = Pattern.compile("^\\s+|\\s+$");
Matcher m = trimmer.matcher(s);
StringBuffer out = new StringBuffer();
while(m.find())
m.appendReplacement(out, "");
m.appendTail(out);
System.out.println(out+"!");
String s="Test ";
s= s.trim();
I prefer not to use regular expressions for trivial problems. This would be a simple option:
public static String trim(final String s) {
final StringBuilder sb = new StringBuilder(s);
while (sb.length() > 0 && Character.isWhitespace(sb.charAt(0)))
sb.deleteCharAt(0); // delete from the beginning
while (sb.length() > 0 && Character.isWhitespace(sb.charAt(sb.length() - 1)))
sb.deleteCharAt(sb.length() - 1); // delete from the end
return sb.toString();
}
Use the String class trim method. It will remove all leading and trailing whitespace.
http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html
String s=" Hello World ";
s = s.trim();
For more information See This
Simply use trim(). It only eliminate the start and end excess white spaces of a string.
String fav = " I like apple ";
fav = fav.trim();
System.out.println(fav);
Output:
I like apple //no extra space at start and end of the string
String.trim() answers the question but was not an option for me.
As stated here :
it simply regards anything up to and including U+0020 (the usual space character) as whitespace, and anything above that as non-whitespace.
This results in it trimming the U+0020 space character and all “control code” characters below U+0020 (including the U+0009 tab character), but not the control codes or Unicode space characters that are above that.
I am working with Japanese where we have full-width characters Ｌｉｋｅ ｔｈｉｓ, the full-width space would not be trimmed by String.trim().
I therefore made a function which, like xehpuk's snippet, use Character.isWhitespace().
However, this version is not using a StringBuilder and instead of deleting characters, finds the 2 indexes it needs to take a trimmed substring out of the original String.
public static String trimWhitespace(final String stringToTrim) {
int endIndex = stringToTrim.length();
// Return the string if it's empty
if (endIndex == 0) return stringToTrim;
int firstIndex = -1;
// Find first character which is not a whitespace, if any
// (increment from beginning until either first non whitespace character or end of string)
while (++firstIndex < endIndex && Character.isWhitespace(stringToTrim.charAt(firstIndex))) { }
// If firstIndex did not reach end of string, Find last character which is not a whitespace,
// (decrement from end until last non whitespace character)
while (--endIndex > firstIndex && Character.isWhitespace(stringToTrim.charAt(endIndex))) { }
// Return substring using indexes
return stringToTrim.substring(firstIndex, endIndex + 1);
}
s = s.trim();
More info:
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#trim()
Why do you not want to use predefined methods? They are usually most efficient.
See String#trim() method
Since Java 11 String class has strip() method which is used to returns a string whose value is this string, with all leading and trailing white space removed. This is introduced to overcome the problem of trim method.
Docs: https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/lang/String.html#strip()
Example:
String str = " abc ";
// public String strip()
str = str.strip(); // Returns abc
There are two more useful methods in Java 11+ String class:
stripLeading() : Returns a string whose value is this string,
with all leading white space removed.
stripTrailing() : Returns a string whose value is this string,
with all trailing white space removed.
While #xehpuk's method is good if you want to avoid using regex, but it has O(n^2) time complexity. The following solution also avoids regex, but is O(n):
if(s.length() == 0)
return "";
char left = s.charAt(0);
char right = s.charAt(s.length() - 1);
int leftWhitespace = 0;
int rightWhitespace = 0;
boolean leftBeforeRight = leftWhitespace < s.length() - 1 - rightWhitespace;
while ((left == ' ' || right == ' ') && leftBeforeRight) {
if(left == ' ') {
leftWhitespace++;
left = s.charAt(leftWhitespace);
}
if(right == ' ') {
rightWhitespace++;
right = s.charAt(s.length() - 1 - rightWhitespace);
}
leftBeforeRight = leftWhitespace < s.length() - 1 - rightWhitespace;
}
String result = s.substring(leftWhitespace, s.length() - rightWhitespace);
return result.equals(" ") ? "" : result;
This counts the number of trailing whitespaces in the beginning and end of the string, until either the "left" and "right" indices obtained from whitespace counts meet, or both indices have reached a non-whitespace character. Afterwards, we either return the substring obtained using the whitespace counts, or the empty string if the result is a whitespace (needed to account for all-whitespace strings with odd number of characters).
I'm writing a small JAVA program which:
takes a text as a String
takes 2 arrays of chars
What im trying to do will sound like "find and replace" but it is not the same so i thought its important to clear it.
Anyway I want to take this text, find if any char from the first array match a char in the text and if so, replace it with the matching char (according to index) from the second char array.
I'll explain with an example:
lets say my text (String) is: "java is awesome!";
i have 2 arrays (char[]): "absm" and "!#*$".
The wished result is to change 'a' to '!' , 'b' to '#' and so on..
meaning the resulted text will be:
"java is awesome!" changed to -> "j#v# i* #w*o$e!"
What is the most efficient way of doing this and why?
I thought about looping the text, but then i found it not so efficient.
(StringBuilder/String class can be used)
StringBuilder sb = new StringBuilder(text);
for(int i = 0; i<text.length(); i ++)
{
for (int j = 0; j < firstCharArray.length;j++)
{
if (sb.charAt(i) == firstCharArray[j])
{
sb.setCharAt(i, secondCharArray[j]);
break;
}
}
}
This way is efficient because it uses a StringBuilder to change the characters in place (if you used Strings you would have to create new ones each time because they are immutable.) Also it minimizes the amount of passes you have to do (1 pass through the text string and n passes through the first array where n = text.length())
I guess you are looking for StringUtils.replaceEach, at least as a reference.
How efficient do you need it to be? Are you doing this for hundreds, thousands, millions of words???
I don't know if it's the most efficent, but you could use the string indexOf() method on each of your possible tokens, it will tell you if it's there, and then you can replace that index at the same time with the corresponding char from the other array.
Codewise, something like (this is half pseudo code by the way):
for(each of first array) {
int temp = YourString.indexOf(current array field);
if (temp >=0) {
replace with other array
}
}
Put the 2 arrays you have in a Map
Map<Character, Character> //or Map of Strings
where the key is "a", "b" etc... and the value is the character you want to substitute with - "#" etc....
Then simply replace the keys in your String with the values.
For small stuff like this, an indexOf() search is likely to be faster than a map, while "avoiding" the inner loop of the accepted answer. Of course, the loop is still there, inside String.indexOf(), but it's likely to be optimized to a fare-thee-well by the JIT-compiler, because it's so heavily used.
static String replaceChars(String source, String from, String to)
{
StringBuilder dest = new StringBuilder(source);
for ( int i = 0; i < source.length(); i++ )
{
int foundAt = from.indexOf(source.charAt(i));
if ( foundAt >= 0 )
dest.setCharAt(i,to.charAt(foundAt));
}
return dest.toString();
}
Update: The Oracle/Sun JIT uses SIMD on at least some processors for indexOf(), making it even faster than one would guess.
Since the only way to know if a character should be replaced is to check it, you (or any util method) have to loop through the whole text, character after the other. You can never achieve better complexity than O(n) (n be the number of characters in the text).
This utility class that replaces a char or a group of chars of a String. It is equivalent to bash tr and perl tr///, aka, transliterate.
/**
* Utility class that replaces chars of a String, aka, transliterate.
*
* It's equivalent to bash 'tr' and perl 'tr///'.
*
*/
public class ReplaceChars {
public static String replace(String string, String from, String to) {
return new String(replace(string.toCharArray(), from.toCharArray(), to.toCharArray()));
}
public static char[] replace(char[] chars, char[] from, char[] to) {
char[] output = chars.clone();
for (int i = 0; i < output.length; i++) {
for (int j = 0; j < from.length; j++) {
if (output[i] == from[j]) {
output[i] = to[j];
break;
}
}
}
return output;
}
/**
* For tests!
*/
public static void main(String[] args) {
// Example from: https://en.wikipedia.org/wiki/Caesar_cipher
String string = "THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG";
String from = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
String to = "XYZABCDEFGHIJKLMNOPQRSTUVW";
System.out.println();
System.out.println("Cesar cypher: " + string);
System.out.println("Result: " + ReplaceChars.replace(string, from, to));
}
}
This is the output:
Cesar cypher: THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG
Result: QEB NRFZH YOLTK CLU GRJMP LSBO QEB IXWV ALD
I'm guessing I'm getting this error because the string is trying to substring a null value. But wouldn't the ".length() > 0" part eliminate that issue?
Here is the Java snippet:
if (itemdescription.length() > 0) {
pstmt2.setString(3, itemdescription.substring(0,38));
}
else {
pstmt2.setString(3, "_");
}
I got this error:
java.lang.StringIndexOutOfBoundsException: String index out of range: 38
at java.lang.String.substring(Unknown Source)
at MASInsert2.itemimport(MASInsert2.java:192)
at MASInsert2.processRequest(MASInsert2.java:125)
at MASInsert2.doGet(MASInsert2.java:219)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:627)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:269)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:188)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:213)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:172)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:117)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:108)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:174)
at org.apache.coyote.http11.Http11AprProcessor.process(Http11AprProcessor.java:835)
at org.apache.coyote.http11.Http11AprProtocol$Http11ConnectionHandler.process(Http11AprProtocol.java:640)
at org.apache.tomcat.util.net.AprEndpoint$Worker.run(AprEndpoint.java:1286)
at java.lang.Thread.run(Unknown Source)
It is a pity that substring is not implemented in a way that handles short strings – like in other languages e.g. Python.
Ok, we cannot change that and have to consider this edge case every time we use substr, instead of if-else clauses I would go for this shorter variant:
myText.substring(0, Math.min(6, myText.length()))
I"m guessing i'm getting this error
because the string is trying to
substring a Null value. But wouldn't
the ".length() > 0" part eliminate
that issue?
No, calling itemdescription.length() when itemdescription is null would not generate a StringIndexOutOfBoundsException, but rather a NullPointerException since you would essentially be trying to call a method on null.
As others have indicated, StringIndexOutOfBoundsException indicates that itemdescription is not at least 38 characters long. You probably want to handle both conditions (I assuming you want to truncate):
final String value;
if (itemdescription == null || itemdescription.length() <= 0) {
value = "_";
} else if (itemdescription.length() <= 38) {
value = itemdescription;
} else {
value = itemdescription.substring(0, 38);
}
pstmt2.setString(3, value);
Might be a good place for a utility function if you do that a lot...
I would recommend apache commons lang. A one-liner takes care of the problem.
pstmt2.setString(3, StringUtils.defaultIfEmpty(
StringUtils.subString(itemdescription,0, 38), "_"));
You really need to check if the string's length is greater to or equal to 38.
Java's substring method fails when you try and get a substring starting at an index which is longer than the string.
An easy alternative is to use Apache Commons StringUtils.substring:
public static String substring(String str, int start)
Gets a substring from the specified String avoiding exceptions.
A negative start position can be used to start n characters from the end of the String.
A null String will return null. An empty ("") String will return "".
StringUtils.substring(null, *) = null
StringUtils.substring("", *) = ""
StringUtils.substring("abc", 0) = "abc"
StringUtils.substring("abc", 2) = "c"
StringUtils.substring("abc", 4) = ""
StringUtils.substring("abc", -2) = "bc"
StringUtils.substring("abc", -4) = "abc"
Parameters:
str - the String to get the substring from, may be null
start - the position to start from, negative means count back from the end of the String by this many characters
Returns:
substring from start position, null if null String input
Note, if you can't use Apache Commons lib for some reason, you could just grab the parts you need from the source
// Substring
//-----------------------------------------------------------------------
/**
* <p>Gets a substring from the specified String avoiding exceptions.</p>
*
* <p>A negative start position can be used to start {#code n}
* characters from the end of the String.</p>
*
* <p>A {#code null} String will return {#code null}.
* An empty ("") String will return "".</p>
*
* <pre>
* StringUtils.substring(null, *) = null
* StringUtils.substring("", *) = ""
* StringUtils.substring("abc", 0) = "abc"
* StringUtils.substring("abc", 2) = "c"
* StringUtils.substring("abc", 4) = ""
* StringUtils.substring("abc", -2) = "bc"
* StringUtils.substring("abc", -4) = "abc"
* </pre>
*
* #param str the String to get the substring from, may be null
* #param start the position to start from, negative means
* count back from the end of the String by this many characters
* #return substring from start position, {#code null} if null String input
*/
public static String substring(final String str, int start) {
if (str == null) {
return null;
}
// handle negatives, which means last n characters
if (start < 0) {
start = str.length() + start; // remember start is negative
}
if (start < 0) {
start = 0;
}
if (start > str.length()) {
return EMPTY;
}
return str.substring(start);
}
substring(0,38) means the String has to be 38 characters or longer. If not, the "String index is out of range".
if (itemdescription != null && itemdescription.length() > 0) {
pstmt2.setString(3, itemdescription.substring(0, Math.min(itemdescription.length(), 38)));
} else {
pstmt2.setString(3, "_");
}
I'm assuming your column is 38 characters in length, so you want to truncate itemdescription to fit within the database. A utility function like the following should do what you want:
/**
* Truncates s to fit within len. If s is null, null is returned.
**/
public String truncate(String s, int len) {
if (s == null) return null;
return s.substring(0, Math.min(len, s.length()));
}
then you just call it like so:
String value = "_";
if (itemdescription != null && itemdescription.length() > 0) {
value = truncate(itemdescription, 38);
}
pstmt2.setString(3, value);
itemdescription is shorter than 38 chars. Which is why the StringOutOfBoundsException is being thrown.
Checking .length() > 0 simply makes sure the String has some not-null value, what you need to do is check that the length is long enough. You could try:
if(itemdescription.length() > 38)
...
You must check the String length. You assume that you can do substring(0,38) as long as String is not null, but you actually need the String to be of at least 38 characters length.
When this is appropriate, I use matches instead of substring.
With substring:
if( myString.substring(1,17).equals("Someting I expect") ) {
// Do stuff
}
// Does NOT work if myString is too short
With matches (must use Regex notation):
if( myString.matches("Someting I expect.*") ) {
// Do stuff
}
// This works with all strings
Should anyone face the same problem.
Do this:
str.substring (...(trim()) ;
Hope it helps somebodies 😎
You get this if itemdescription is shorter than 38 characters
You can look which exceptions are thrown and when in the JAVA API
in you case for String#substring(int,int): https://docs.oracle.com/javase/9/docs/api/java/lang/String.html#substring-int-int-
substring
public String substring(int beginIndex, int endIndex)
. . .
Throws:
IndexOutOfBoundsException
if the beginIndex is negative,
or endIndex is larger than the length of this String object,
or beginIndex is larger than endIndex.
(same applies to previous java versions as well)
In Java is there a way to check the condition:
"Does this single character appear at all in string x"
without using a loop?
You can use string.indexOf('a').
If the char a is present in string :
it returns the the index of the first occurrence of the character in
the character sequence represented by this object, or -1 if the
character does not occur.
String.contains() which checks if the string contains a specified sequence of char values
String.indexOf() which returns the index within the string of the first occurence of the specified character or substring (there are 4 variations of this method)
I'm not sure what the original poster is asking exactly. Since indexOf(...) and contains(...) both probably use loops internally, perhaps he's looking to see if this is possible at all without a loop? I can think of two ways off hand, one would of course be recurrsion:
public boolean containsChar(String s, char search) {
if (s.length() == 0)
return false;
else
return s.charAt(0) == search || containsChar(s.substring(1), search);
}
The other is far less elegant, but completeness...:
/**
* Works for strings of up to 5 characters
*/
public boolean containsChar(String s, char search) {
if (s.length() > 5) throw IllegalArgumentException();
try {
if (s.charAt(0) == search) return true;
if (s.charAt(1) == search) return true;
if (s.charAt(2) == search) return true;
if (s.charAt(3) == search) return true;
if (s.charAt(4) == search) return true;
} catch (IndexOutOfBoundsException e) {
// this should never happen...
return false;
}
return false;
}
The number of lines grow as you need to support longer and longer strings of course. But there are no loops/recurrsions at all. You can even remove the length check if you're concerned that that length() uses a loop.
You can use 2 methods from the String class.
String.contains() which checks if the string contains a specified sequence of char values
String.indexOf() which returns the index within the string of the first occurence of the specified character or substring or returns -1 if the character is not found (there are 4 variations of this method)
Method 1:
String myString = "foobar";
if (myString.contains("x") {
// Do something.
}
Method 2:
String myString = "foobar";
if (myString.indexOf("x") >= 0 {
// Do something.
}
Links by: Zach Scrivena
String temp = "abcdefghi";
if(temp.indexOf("b")!=-1)
{
System.out.println("there is 'b' in temp string");
}
else
{
System.out.println("there is no 'b' in temp string");
}
If you need to check the same string often you can calculate the character occurrences up-front. This is an implementation that uses a bit array contained into a long array:
public class FastCharacterInStringChecker implements Serializable {
private static final long serialVersionUID = 1L;
private final long[] l = new long[1024]; // 65536 / 64 = 1024
public FastCharacterInStringChecker(final String string) {
for (final char c: string.toCharArray()) {
final int index = c >> 6;
final int value = c - (index << 6);
l[index] |= 1L << value;
}
}
public boolean contains(final char c) {
final int index = c >> 6; // c / 64
final int value = c - (index << 6); // c - (index * 64)
return (l[index] & (1L << value)) != 0;
}}
To check if something does not exist in a string, you at least need to look at each character in a string. So even if you don't explicitly use a loop, it'll have the same efficiency. That being said, you can try using str.contains(""+char).
Is the below what you were looking for?
int index = string.indexOf(character);
return index != -1;
Yes, using the indexOf() method on the string class. See the API documentation for this method
String.contains(String) or String.indexOf(String) - suggested
"abc".contains("Z"); // false - correct
"zzzz".contains("Z"); // false - correct
"Z".contains("Z"); // true - correct
"😀and😀".contains("😀"); // true - correct
"😀and😀".contains("😂"); // false - correct
"😀and😀".indexOf("😀"); // 0 - correct
"😀and😀".indexOf("😂"); // -1 - correct
String.indexOf(int) and carefully considered String.indexOf(char) with char to int widening
"😀and😀".indexOf("😀".charAt(0)); // 0 though incorrect usage has correct output due to portion of correct data
"😀and😀".indexOf("😂".charAt(0)); // 0 -- incorrect usage and ambiguous result
"😀and😀".indexOf("😂".codePointAt(0)); // -1 -- correct usage and correct output
The discussions around character is ambiguous in Java world
can the value of char or Character considered as single character?
No. In the context of unicode characters, char or Character can sometimes be part of a single character and should not be treated as a complete single character logically.
if not, what should be considered as single character (logically)?
Any system supporting character encodings for Unicode characters should consider unicode's codepoint as single character.
So Java should do that very clear & loud rather than exposing too much of internal implementation details to users.
String class is bad at abstraction (though it requires confusingly good amount of understanding of its encapsulations to understand the abstraction 😒😒😒 and hence an anti-pattern).
How is it different from general char usage?
char can be only be mapped to a character in Basic Multilingual Plane.
Only codePoint - int can cover the complete range of Unicode characters.
Why is this difference?
char is internally treated as 16-bit unsigned value and could not represent all the unicode characters using UTF-16 internal representation using only 2-bytes. Sometimes, values in a 16-bit range have to be combined with another 16-bit value to correctly define character.
Without getting too verbose, the usage of indexOf, charAt, length and such methods should be more explicit. Sincerely hoping Java will add new UnicodeString and UnicodeCharacter classes with clearly defined abstractions.
Reason to prefer contains and not indexOf(int)
Practically there are many code flows that treat a logical character as char in java.
In Unicode context, char is not sufficient
Though the indexOf takes in an int, char to int conversion masks this from the user and user might do something like str.indexOf(someotherstr.charAt(0))(unless the user is aware of the exact context)
So, treating everything as CharSequence (aka String) is better
public static void main(String[] args) {
System.out.println("😀and😀".indexOf("😀".charAt(0))); // 0 though incorrect usage has correct output due to portion of correct data
System.out.println("😀and😀".indexOf("😂".charAt(0))); // 0 -- incorrect usage and ambiguous result
System.out.println("😀and😀".indexOf("😂".codePointAt(0))); // -1 -- correct usage and correct output
System.out.println("😀and😀".contains("😀")); // true - correct
System.out.println("😀and😀".contains("😂")); // false - correct
}
Semantics
char can handle most of the practical use cases. Still its better to use codepoints within programming environment for future extensibility.
codepoint should handle nearly all of the technical use cases around encodings.
Still, Grapheme Clusters falls out of the scope of codepoint level of abstraction.
Storage layers can choose char interface if ints are too costly(doubled). Unless storage cost is the only metric, its still better to use codepoint. Also, its better to treat storage as byte and delegate semantics to business logic built around storage.
Semantics can be abstracted at multiple levels. codepoint should become lowest level of interface and other semantics can be built around codepoint in runtime environment.
package com;
public class _index {
public static void main(String[] args) {
String s1="be proud to be an indian";
char ch=s1.charAt(s1.indexOf('e'));
int count = 0;
for(int i=0;i<s1.length();i++) {
if(s1.charAt(i)=='e'){
System.out.println("number of E:=="+ch);
count++;
}
}
System.out.println("Total count of E:=="+count);
}
}
static String removeOccurences(String a, String b)
{
StringBuilder s2 = new StringBuilder(a);
for(int i=0;i<b.length();i++){
char ch = b.charAt(i);
System.out.println(ch+" first index"+a.indexOf(ch));
int lastind = a.lastIndexOf(ch);
for(int k=new String(s2).indexOf(ch);k > 0;k=new String(s2).indexOf(ch)){
if(s2.charAt(k) == ch){
s2.deleteCharAt(k);
System.out.println("val of s2 : "+s2.toString());
}
}
}
System.out.println(s1.toString());
return (s1.toString());
}
you can use this code. It will check the char is present or not. If it is present then the return value is >= 0 otherwise it's -1. Here I am printing alphabets that is not present in the input.
import java.util.Scanner;
public class Test {
public static void letters()
{
System.out.println("Enter input char");
Scanner sc = new Scanner(System.in);
String input = sc.next();
System.out.println("Output : ");
for (char alphabet = 'A'; alphabet <= 'Z'; alphabet++) {
if(input.toUpperCase().indexOf(alphabet) < 0)
System.out.print(alphabet + " ");
}
}
public static void main(String[] args) {
letters();
}
}
//Ouput Example
Enter input char
nandu
Output :
B C E F G H I J K L M O P Q R S T V W X Y Z
If you see the source code of indexOf in JAVA:
public int indexOf(int ch, int fromIndex) {
final int max = value.length;
if (fromIndex < 0) {
fromIndex = 0;
} else if (fromIndex >= max) {
// Note: fromIndex might be near -1>>>1.
return -1;
}
if (ch < Character.MIN_SUPPLEMENTARY_CODE_POINT) {
// handle most cases here (ch is a BMP code point or a
// negative value (invalid code point))
final char[] value = this.value;
for (int i = fromIndex; i < max; i++) {
if (value[i] == ch) {
return i;
}
}
return -1;
} else {
return indexOfSupplementary(ch, fromIndex);
}
}
you can see it uses a for loop for finding a character. Note that each indexOf you may use in your code, is equal to one loop.
So, it is unavoidable to use loop for a single character.
However, if you want to find a special string with more different forms, use useful libraries such as util.regex, it deploys stronger algorithm to match a character or a string pattern with Regular Expressions. For example to find an email in a string:
String regex = "^(.+)#(.+)$";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(email);
If you don't like to use regex, just use a loop and charAt and try to cover all cases in one loop.
Be careful recursive methods has more overhead than loop, so it's not recommended.
how about one uses this ;
let text = "Hello world, welcome to the universe.";
let result = text.includes("world");
console.log(result) ....// true
the result will be a true or false
this always works for me
You won't be able to check if char appears at all in some string without atleast going over the string once using loop / recursion ( the built-in methods like indexOf also use a loop )
If the no. of times you look up if a char is in string x is more way more than the length of the string than I would recommend using a Set data structure as that would be more efficient than simply using indexOf
String s = "abc";
// Build a set so we can check if character exists in constant time O(1)
Set<Character> set = new HashSet<>();
int len = s.length();
for(int i = 0; i < len; i++) set.add(s.charAt(i));
// Now we can check without the need of a loop
// contains method of set doesn't use a loop unlike string's contains method
set.contains('a') // true
set.contains('z') // false
Using set you will be able to check if character exists in a string in constant time O(1) but you will also use additional memory ( Space complexity will be O(n) ).