Java解析字符串为长整形

把变通数字串、带小数点形式和带E形式的字符串转化为长整形，考虑可能欠全面，请大家指点。

/**
	 * 字符串转化为long,不考虑浮点数可能出现的四舍五入问题
	 * @param str 字符串的格式："123"\"23.23"\".12"\"23."\"23.3E3"\".2E3"\
	 * @return
	 */
	public long string2Long(String str){
		long result = 0;
		
		String regex1 = "\\d+";
		String regex2 = "\\d*\\.\\d*";
		String regex3 = "\\d*[.]?\\d+[eE]\\d*";
		
		if(Pattern.matches(regex1, str)){
			result = Long.parseLong(str);
		}else if(Pattern.matches(regex2, str)){
			int index = str.indexOf(".");
			if(index == 0){
				result = 0;
			}else{
				str = str.substring(0, index);
				result = Long.parseLong(str);
			}
		}else if(Pattern.matches(regex3, str)){
			int index = 0;
			String exp = "0";
			if(str.contains("e")){
				index = str.indexOf("e");
				str = str.substring(0, index);
				exp = str.substring(index + 1);
			}else if(str.contains("E")){
				index = str.indexOf("E");
				if(index < str.length()){
					exp = str.substring(index + 1);
				}
				str = str.substring(0, index);
			}
			if(str == null){
				result = 0;
			}else{
				double pre = Double.parseDouble(str);
				long lexp = Long.parseLong(exp);
				result = new Double(pre * Math.pow(10, lexp)).longValue();
			}
		}else{
			throw new RuntimeException("格式错误");
		}
		
		return result;
	}