Java - 检查STRING是否只包含某些字符的最佳方法是什么？

Question

我有这个问题:我有一个String,但我需要确保它只包含字母AZ和数字0-9.这是我目前的代码:

boolean valid = true;
for (char c : string.toCharArray()) {
    int type = Character.getType(c);
    if (type == 2 || type == 1 || type == 9) {
        // the character is either a letter or a digit
    } else {
        valid = false;
        break;
    }
}

但实施它的最佳和最有效的方法是什么？

Answer 1

由于没有其他人担心"最快",这是我的贡献:

boolean valid = true;

char[] a = s.toCharArray();

for (char c: a)
{
    valid = ((c >= 'a') && (c <= 'z')) || 
            ((c >= 'A') && (c <= 'Z')) || 
            ((c >= '0') && (c <= '9'));

    if (!valid)
    {
        break;
    }
}

return valid;

完整测试代码如下:

public static void main(String[] args)
{
    String[] testStrings = {"abcdefghijklmnopqrstuvwxyz0123456789", "", "00000", "abcdefghijklmnopqrstuvwxyz0123456789&", "1", "q", "test123", "(#*$))&v", "ABC123", "hello", "supercalifragilisticexpialidocious"};

    long startNanos = System.nanoTime();

    for (String testString: testStrings)
    {
        isAlphaNumericOriginal(testString);
    }

    System.out.println("Time for isAlphaNumericOriginal: " + (System.nanoTime() - startNanos) + " ns"); 

    startNanos = System.nanoTime();

    for (String testString: testStrings)
    {
        isAlphaNumericFast(testString);
    }

    System.out.println("Time for isAlphaNumericFast: " + (System.nanoTime() - startNanos) + " ns");

    startNanos = System.nanoTime();

    for (String testString: testStrings)
    {
        isAlphaNumericRegEx(testString);
    }

    System.out.println("Time for isAlphaNumericRegEx: " + (System.nanoTime() - startNanos) + " ns");

    startNanos = System.nanoTime();

    for (String testString: testStrings)
    {
        isAlphaNumericIsLetterOrDigit(testString);
    }

    System.out.println("Time for isAlphaNumericIsLetterOrDigit: " + (System.nanoTime() - startNanos) + " ns");      
}

private static boolean isAlphaNumericOriginal(String s)
{
    boolean valid = true;
    for (char c : s.toCharArray()) 
    {
        int type = Character.getType(c);
        if (type == 2 || type == 1 || type == 9) 
        {
            // the character is either a letter or a digit
        }
        else 
        {
            valid = false;
            break;
        }
    }

    return valid;
}

private static boolean isAlphaNumericFast(String s)
{
    boolean valid = true;

    char[] a = s.toCharArray();

    for (char c: a)
    {
        valid = ((c >= 'a') && (c <= 'z')) || 
                ((c >= 'A') && (c <= 'Z')) || 
                ((c >= '0') && (c <= '9'));

        if (!valid)
        {
            break;
        }
    }

    return valid;
}

private static boolean isAlphaNumericRegEx(String s)
{
    return Pattern.matches("[\\dA-Za-z]+", s);
}

private static boolean isAlphaNumericIsLetterOrDigit(String s)
{
    boolean valid = true;
    for (char c : s.toCharArray()) { 
        if(!Character.isLetterOrDigit(c))
        {
            valid = false;
            break;
        }
    }
    return valid;
}

为我生成此输出:

Time for isAlphaNumericOriginal: 164960 ns
Time for isAlphaNumericFast: 18472 ns
Time for isAlphaNumericRegEx: 1978230 ns
Time for isAlphaNumericIsLetterOrDigit: 110315 ns

Answer 2

如果你想避免正则表达式,那么这个Character类可以帮助:

boolean valid = true;
for (char c : string.toCharArray()) { 
    if(!Character.isLetterOrDigit(c))
    {
        valid = false;
        break;
    }
}

如果你关心的是大写,那么请在if if语句下面做:

if(!((Character.isLetter(c) && Character.isUpperCase(c)) || Character.isDigit(c)))