检查字符串的最快方法是Java中的字母数字 [英] Fastest way to check a string is alphanumeric in Java

问题描述

检查字符串是否只包含字母数字字符的最快方法是什么。

What is the fastest way to check that a String contains only alphanumeric characters.

我有一些代码会嚼掉很多CPU和我想知道是否会比使用预编译的正则表达式更快捷。

I've got some code that is going to chew up a lot of CPU and I wonder if there is going to be a quicker way than using pre-compiled regular expressions.

更新：这么多的投票，关心说明？不应该使用Stackoverflow来讨论用于实现任务的算法可能更快？

UPDATE: So many down votes, care to explain? Stackoverflow should not be used to discuss which algorithm to use to achieve a task might be faster?

推荐答案

我已经编写了测试验证正确的答案。在运行Java 1.6的四核OSX10.8计算机上进行的测试

I've written the tests that verify the "correct" answer. Tests done on a quad core OSX10.8 machine running Java 1.6

有趣的是，使用正则表达式比手动迭代字符串慢大约5-10倍。此外， isAlphanumeric2（）函数略快于 isAlphanumeric（）。

Interestingly using regular expressions turns out to be about 5-10 times slower than manually iterating over a string. Furthermore the isAlphanumeric2() function is marginally faster than isAlphanumeric().

public class QuickTest extends TestCase {

    private final int reps = 1000000;

    public void testRegexp() {
        for(int i = 0; i < reps; i++)
            ("ab4r3rgf"+i).matches("[a-zA-Z0-9]");
    }

public void testIsAlphanumeric() {
    for(int i = 0; i < reps; i++)
        isAlphanumeric("ab4r3rgf"+i);
}

public void testIsAlphanumeric2() {
    for(int i = 0; i < reps; i++)
        isAlphanumeric2("ab4r3rgf"+i);
}

    public boolean isAlphanumeric(String str) {
        for (int i=0; i<str.length(); i++) {
            char c = str.charAt(i);
            if (!Character.isDigit(c) && !Character.isLetter(c))
                return false;
        }

        return true;
    }

    public boolean isAlphanumeric2(String str) {
        for (int i=0; i<str.length(); i++) {
            char c = str.charAt(i);
            if (c < 0x30 || (c >= 0x3a && c <= 0x40) || (c > 0x5a && c <= 0x60) || c > 0x7a)
                return false;
        }
        return true;
    }

}

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