Valid Number

Validate if a given string can be interpreted as a decimal number.

Example

Some examples: "0"=>true " 0.1 "=>true "abc"=>false "1 a"=>false "2e10"=>true " -90e3 "=>true " 1e"=>false "e3"=>false " 6e-1"=>true " 99e2.5 "=>false "53.5e93"=>true " --6 "=>false "-+3"=>false "95a54e53"=>false

Note:It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:

• Numbers 0-9

• Exponent - "e"

• Positive/negative sign - "+"/"-"

• Decimal point - "."

Of course, the context of these characters also matters in the input.

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

Note

计算point和digit的出现次数，point多于1或者digit少于1都不行

对于自然对数，最后再次判断

Code

class Solution {
    public boolean isNumber(String s) {
        int nPiont = 0, nDigit = 0;
        s = s.trim() + " ";
        char[] c = s.toCharArray();
        int i = 0;
        int len = c.length - 1;
        if (c[i] == '+' || c[i] == '-') {
            i++;
        }
        while (Character.isDigit(c[i]) || c[i] == '.') {
            if (Character.isDigit(c[i])) {
                nDigit++;
            }
            if (c[i] == '.') {
                nPiont++;
            }
            i++;
        }
        if (nPiont > 1) {
            return false;
        }
        if (nDigit < 1) {
            return false;
        }
        if (c[i] == 'e') {
            i++;
            if (c[i] == '+' || c[i] == '-') {
                i++;
            }
            if (i == len) {
                return false;
            }
            for (; i < len; i++) {
                if (!Character.isDigit(c[i])) {
                    return false;
                }
            }
        }
        return i == len;
    }
}