*Valid Number

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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

解法一：

/**
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 */

// Non-regex version

public class Solution {
    public boolean isNumber(String s) {
        int len = s.length();
        int i = 0, e = len - 1;
        while (i <= e && Character.isWhitespace(s.charAt(i))) i++;
        if (i > len - 1) return false;
        while (e >= i && Character.isWhitespace(s.charAt(e))) e--;
        // skip leading +/-
        if (s.charAt(i) == '+' || s.charAt(i) == '-') i++;
        boolean num = false; // is a digit
        boolean dot = false; // is a '.'
        boolean exp = false; // is a 'e'
        while (i <= e) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) {
                num = true;
            }
            else if (c == '.') {
                if(exp || dot) return false;
                dot = true;
            }
            else if (c == 'e') {
                if(exp || num == false) return false;
                exp = true;
                num = false;
            }
            else if (c == '+' || c == '-') {
                if (s.charAt(i - 1) != 'e') return false;
            }
            else {
                return false;
            }
            i++;
        }
        return num;
    }
}

解法二：

CleanCode Version: better logic

public class Solution {
    public boolean isNumber(String s) 
    {
        int i=0; 
        int n=s.length();
        while(i<n&&Character.isWhitespace(s.charAt(i)))i++;
        if(i<n&&(s.charAt(i)=='+'||s.charAt(i)=='-'))i++;
        boolean isNumeric = false;
        while(i<n&&Character.isDigit(s.charAt(i)))
        {
            i++;
            isNumeric = true;
        }
        if(i<n&&s.charAt(i)=='.')
        {
            i++;
            while(i<n&&Character.isDigit(s.charAt(i)))
            {
                i++;
                isNumeric =  true;
            }
        }
        if(isNumeric&&i<n&&s.charAt(i)=='e')
        {
            i++;
            isNumeric = false;
            if(i<n&&(s.charAt(i)=='+'||s.charAt(i)=='-'))i++;
            while(i<n&&Character.isDigit(s.charAt(i)))
            {
                i++;
                isNumeric =  true;
            }
        }
        while(i<n&&Character.isWhitespace(s.charAt(i)))i++;
        return i==n&&isNumeric;
    }
}