第28题 Valid Number

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Validate if a given string is numeric.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

Solution in Java：

public class Solution {    public boolean isNumber(String s) {        String sTrim = s.trim();        if(sTrim.length()==0) return false;        if(sTrim.contains(" ")) return false;                String pattern1 = "", pattern2 = "";        if(sTrim.contains("e")){            int indexE = sTrim.indexOf('e');            String beforeE="", afterE="";            if(indexE!=-1){                beforeE = sTrim.substring(0,indexE);                afterE = sTrim.substring(indexE+1, sTrim.length());                indexE = sTrim.indexOf('e', indexE+1);                if(indexE!=-1) return false;            }            pattern1 = "^[+,-]?[0-9]+$";            if(!afterE.matches(pattern1))                 return false;            if(beforeE.contains(".")){                pattern1 = "^[+,-]?[0-9]*[.]{1}[0-9]+$";                pattern2 = "^[+,-]?[0-9]+[.]{1}[0-9]*$";                if(beforeE.matches(pattern1)||beforeE.matches(pattern2)) return true;                else return false;            }else{                pattern1 = "^[+,-]?[0-9]+$";                if(beforeE.matches(pattern1)) return true;                else return false;            }        }        else{            if(sTrim.contains(".")){                pattern1 = "^[+,-]?[0-9]+[.]{1}[0-9]*$";                pattern2 = "^[+,-]?[0-9]*[.]{1}[0-9]+$";                if(sTrim.matches(pattern1)||sTrim.matches(pattern2)) return true;                else return false;            }            else{                pattern1 = "^[+,-]?[0-9]+$";                if(sTrim.matches(pattern1)) return true;                else return false;            }    }}

Note：

“.3”也算valid number！"3."也算valid number！但"."前后不可同时为空字符，所以要用或判断。

"7e69e"在java中用split分割，长度为2，最后不算第二个e后的空字符，所以不能检查如“e123e”格式的字符串。所以用indexOf函数查找字符"e"，然后用substring获取"e"前后字符。string.indexOf('e',index)查找从string的index位置开始有无字符'e'。

科学计数法e后可以加＋或－号，后面跟数字。

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