我知道有一百万种方式做到这一点,但最快的是什么？这应该包括科学记数法.
注意：我没有兴趣将值转换为Double,我只想知道是否可能.即private boolean isDouble(String value).
解决方法
您可以使用Double类使用的相同正则表达式来检查它.这里有很好的记录：
http://docs.oracle.com/javase/6/docs/api/java/lang/Double.html#valueOf%28java.lang.String%29
这是代码部分：
To avoid calling this method on an invalid string and having a NumberFormatException be thrown,the regular expression below can be used to screen the input string:
final String Digits = "(\\p{Digit}+)";
final String HexDigits = "(\\p{XDigit}+)";
// an exponent is 'e' or 'E' followed by an optionally
// signed decimal integer.
final String Exp = "[eE][+-]?"+Digits;
final String fpRegex =
("[\\x00-\\x20]*"+ // Optional leading "whitespace"
"[+-]?(" + // Optional sign character
"NaN|" + // "NaN" string
"Infinity|" + // "Infinity" string
// A decimal floating-point string representing a finite positive
// number without a leading sign has at most five basic pieces:
// Digits . Digits ExponentPart FloatTypeSuffix
//
// Since this method allows integer-only strings as input
// in addition to strings of floating-point literals,the
// two sub-patterns below are simplifications of the grammar
// productions from the Java Language Specification,2nd
// edition,section 3.10.2.
// Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt
"((("+Digits+"(\\.)?("+Digits+"?)("+Exp+")?)|"+
// . Digits ExponentPart_opt FloatTypeSuffix_opt
"(\\.("+Digits+")("+Exp+")?)|"+
// Hexadecimal strings
"((" +
// 0[xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt
"(0[xX]" + HexDigits + "(\\.)?)|" +
// 0[xX] HexDigits_opt . HexDigits BinaryExponent FloatTypeSuffix_opt
"(0[xX]" + HexDigits + "?(\\.)" + HexDigits + ")" +
")[pP][+-]?" + Digits + "))" +
"[fFdD]?))" +
"[\\x00-\\x20]*");// Optional trailing "whitespace"
if (Pattern.matches(fpRegex,myString))
Double.valueOf(myString); // Will not throw NumberFormatException
else {
// Perform suitable alternative action
}
