How Can I Efficiently Replace Multiple Substrings in a Java String?
Nov 05, 2024 am 06:42 AMReplacing Multiple Substrings in a String Efficiently in Java
When confronted with the need to replace multiple substrings within a string, it is tempting to resort to the brute force approach of repeatedly applying the string.replace() method. However, this can be inefficient for large strings or when working with numerous strings.
Exploiting Regular Expressions
A more efficient solution involves leveraging regular expressions. Regular expressions allow you to define complex search patterns and perform text transformations in a single operation.
Example Usage
Consider a scenario where you want to replace tokens such as "?t%" and "?verage%" with values stored in a map. Using StringUtils from Apache Commons Lang, you can create a pattern and matcher as follows:
<code class="java">Map<String,String> tokens = new HashMap<>(); tokens.put("cat", "Garfield"); tokens.put("beverage", "coffee"); String template = "%cat% really needs some %beverage%."; // Create pattern of the format "%(cat|beverage)%" String patternString = "%(" + StringUtils.join(tokens.keySet(), "|") + ")%"; Pattern pattern = Pattern.compile(patternString); Matcher matcher = pattern.matcher(template);</code>
Next, you can use the matcher to find all occurrences of the tokens and replace them with the corresponding values:
<code class="java">StringBuffer sb = new StringBuffer(); while(matcher.find()) { matcher.appendReplacement(sb, tokens.get(matcher.group(1))); } matcher.appendTail(sb); System.out.println(sb.toString());</code>
Benefits of Regular Expressions
Once the regular expression is compiled, searching the input string is typically very fast. Additionally, regular expressions provide the flexibility to handle complex search patterns, such as those involving parentheses and quantifiers.
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