Given a
String text = "RHKKA";
How to efficiently replace all 'R' with 'H', all 'H' with 'E', all 'K' with 'L' and all 'A' with 'O'?
String text would be HELLO then.
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2Do you have mappings for an entire alphabet, or just those letters?Elliott Frisch– Elliott Frisch2018-08-26 17:32:47 +00:00Commented Aug 26, 2018 at 17:32
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4 Answers
You can create a Map of Character as key and value, then loop over character by character like so :
String text = "RHKKA";
Map<Character, Character> map = new HashMap<>();
map.put('R', 'H');
map.put('H', 'E');
map.put('K', 'L');
map.put('A', 'O');
char[] chars = text.toCharArray();
for (int i = 0; i < chars.length; i++) {
chars[i] = map.get(chars[i]);
}
String result = String.valueOf(chars);
System.out.println(result.toString());//HELLO
Java8+ possible solution
Or if you are using Java8+ you can use :
String result = text.chars()
.mapToObj(c -> String.valueOf(map.get((char) c)))
.collect(Collectors.joining());//HELLO
Java9+ possible solution
Another possible solution if to use Matcher::replaceAll like so :
String text = "RHKKA";
Map<Character, Character> map = Map.of('R', 'H', 'H', 'E', 'K', 'L', 'A', 'O');
text = Pattern.compile(".").matcher(text)
.replaceAll(c -> String.valueOf(map.get(c.group().charAt(0))));//HELLO
You can read more about Map.of
6 Comments
Andy Turner
Of course, you need to ensure the mapping is complete (enough), otherwise your output string would be peppered with
nulls.Andy Turner
Also perhaps worth pointing out that you don't need the
StringBuilder in the first way, you can just update the char[] from toCharArray(), and then build a new string from that. (Or construct the StringBuilder with the entire text, and update it one char at a time. Kinda the same approach.)Andy Turner
@YCF_L yes. And for the second part, I meant that you could do largely the same with
StringBuilder result = new StringBuilder(text) and charAt/set, since a StringBuilder is just a wrapper around a char[].
Youcef LAIDANI
Thank you @AndyTurner good idea I edit my answer with this one
GhostCat
I wish I could upvote twice, for the Java9 Map.of() ;-)
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There is no such function in standard Java but you can look at Apache Commons StringUtils.replaceChars which replaces group with another group in one go, effectively doing the same as Unix tr command:
StringUtils.replaceChars("RHKKA", "RHKA", "HELE") = "HELLO".
answered Aug 26, 2018 at 17:33
Zbynek Vyskovsky - kvr000
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Comments
I think your best bet is to create a new String or StringBuilder by replacing all characters in the original one with the correct ones.
Something like:
String text = "RHKKA";
StringBuilder newString = new StringBuilder(text.length());
for (int i = 0; i < text.length(); i++) {
char newChar = text.charAt(i);
if (text.charAt(i) == 'R')
newChar = 'H';
else if (text.charAt(i) == 'H')
newChar = 'E';
else if (text.charAt(i) == 'K')
newChar = 'L';
else if (text.charAt(i) == 'A')
newChar = 'O';
newString.append(newChar);
}
System.out.println(newString); //prints HELLO
Comments
Here is an alternative way based on the character replacement:
String text = "RHKKA";
String before = "RHKA"; // before mapping
String after = "HELO"; // after mapping
String output = Arrays.stream(text.split("")) // Split by characters
.map(i -> String.valueOf(after.charAt(before.indexOf(i)))) // Replace each letter
.collect(Collectors.joining("")); // Collect to String
System.out.println(output); // HELLO
By the way, I am sure you meant to replace A with O.. not E as you have in the description.
