子串或替换更快以删除字符串中的最后一个字符？

Question

我有很多需要处理的单词并且所有单词都以此结尾.,哪个选项具有最佳的时间复杂度？

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

2. word.replaceAll("\\."?"")

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

或者,还有更好的方法？

Answer 1

一个简单的测试(使用JDK1.7.0_75)可以​​显示差异:

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.");

}

输出:

子串:0毫秒.

replaceAll:78毫秒.

替换:16毫秒.

正如所料,这substring是最快的,因为:

1. 它避免编译正则表达式.
2. 它是恒定时间:String基于指定的开始和结束索引创建新的.