如何有效替换字符串中的字符

Question

I have a string like this:

String line="[x1,x2,x3]";

I want to remove both of the square brackets from the String (replace them with an empty string). I tried it like this:

 String x=line.replace("[","").replace("]","");

Is there a more efficient way than this?

Answer 1

You may use .replaceAll if you want to use a regex to remove both [ and ] at the same time anywhere inside the string (actually, the equivalent of OP two line code snippet):

String x = line.replaceAll("[\\]\\[]+", "");

The [\\]\\[]+ pattern matches one or more ( + ) [ or ] characters (these must be escaped inside [....] , a character class).

Java demo :

String line="[x1,x2,x3]";
String x = line.replaceAll("[\\[\\]]+", "");
System.out.println(x);
// => x1,x2,x3

Answer 2

String is immutable so you can

String x = line.replace("[","").replace("]","");

another more efficient way is using regex with a pattern for both [ and ]

like

String x = line.replaceAll("[\\]\\[]+", "");

Answer 3

如果方括号始终存在，并且始终如示例所示在字符串的开头和结尾，则可以只使用substring ：

String x2 = line.substring(1, line.length()-1);

Answer 4

Another way:

line = line.substring(1, line.length()-1);

---

If you want to preserve the square brackets that comes in between, then you can use:

String line="[x1,[x2],x3]";
line = line.replaceAll("^\\[(.*)\\]$", "$1");

Output

x1,[x2],x3

Answer 5

If brackets can only appear at the beginning and at the end of the string, the fastest method (in terms of performance) would be using substring which does not involve regular expressions.

If brackets are guaranteed to be present:

line = line.substring(1, line.length()-1);

If brackets are optional:

int len = line.length();
int trimStart = (len > 0 && line.charAt(0) == '[') ? 1 : 0;
int trimEnd = (len > 0 && line.charAt(len-1) == ']') ? 1 : 0;
if (trimStart > 0 || trimEnd > 0) {
    line = line.substring(trimStart, len - trimEnd);
}