子串或替换更快以删除字符串中的最后一个字符？

Question

I have lots of words that need processing and all of them end with . , Which option has the best time complexity?

1. word.substring(0, word.length()-1)

2. word.replaceAll("\\\\."，"")

3. word.replace(".", "")

Or, is there a better way?

Answer 1

A simple test (with JDK1.7.0_75) can show the difference:

private static final int LENGTH = 10000;

public static void main(String[] args) {
    String[] strings = new String[LENGTH];
    for (int i = 0; i < LENGTH; i++) {
        strings[i] = "abc" + i + ".";
    }
    long start = System.currentTimeMillis();
    for (int i = 0; i < strings.length; i++) {
        String word = strings[i];
        word = word.substring(0, word.length()-1);
    }
    long end = System.currentTimeMillis();

    System.out.println("substring: " + (end - start) + " millisec.");

    start = System.currentTimeMillis();
    for (int i = 0; i < strings.length; i++) {
        String word = strings[i];
        word = word.replaceAll(".", "");
    }
    end = System.currentTimeMillis();

    System.out.println("replaceAll: " + (end - start) + " millisec.");

    start = System.currentTimeMillis();
    for (int i = 0; i < strings.length; i++) {
        String word = strings[i];
        word = word.replace(".", "");
    }
    end = System.currentTimeMillis();

    System.out.println("replace: " + (end - start) + " millisec.");

}

Output:

substring: 0 millisec.

replaceAll: 78 millisec.

replace: 16 millisec.

As expected, the substring is fastest because:

1. It avoids compiling a regular expression.
2. It is constant-time: creating a new String based on the specified begin and end indices.