I want to replace all the characters in a Java String with *
character. So it shouldn\'t matter what character it is, it should be replaced with a *
public String allStar(String s) {
StringBuilder sb = new StringBuilder(s.length());
for (int i = 0; i < s.length(); i++) {
sb.append('*');
}
return sb.toString();
}
String nCopies(String s, int n) {
return n == 1 ? s.replaceFirst(".$", "") : nCopies(s + s, --n);
}
String text = "Hello World";
System.out.println( text.replaceAll( "[A-Za-z0-9]", "*" ) );
output : ***** *****
There may be other faster/better ways to do it, but you could just use a string buffer and a for-loop:
public String stringToAsterisk(String input) {
if (input == null) return "";
StringBuffer sb = new StringBuffer();
for (int x = 0; x < input.length(); x++) {
sb.append("*");
}
return sb.toString();
}
If your application is single threaded, you can use StringBuilder instead, but it's not thread safe.
I am not sure if this might be any faster:
public String stringToAsterisk(String input) {
if (input == null) return "";
int length = input.length();
char[] chars = new char[length];
while (length > 0) chars[--length] = "*";
return new String(chars);
}
str = "*".repeat(str.length());
Note: This replaces newlines \n
with *
. If you want to preserve \n
, see solution below.
str = str.replaceAll(".", "*");
This preserves newlines.
To replace newlines with *
as well in Java 10 and earlier, you can use:
str = str.replaceAll("(?s).", "*");
The (?s)
doesn't match anything but activates DOTALL mode which makes .
also match \n
.
Without any external library and without your own loop, you can do:
String input = "Hello";
char[] ca = new char[input.length()];
Arrays.fill(ca, '*');
String output = new String(ca);
BTW, both Arrays.fill()
and String(char [])
are really fast.