Valid Number(Hard)

Validate if a given string is numeric.

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

Logic:

1. If we see [0-9] we reset the number flags.
2. We can only see . if we didn't see e or ..
3. We can only see e if we didn't see e but we did see a number. We reset numberAfterE flag.
4. We can only see + and - in the beginning and after an e
5. any other character break the validation.

At the and it is only valid if there was at least 1 number and if we did see an e then a number after it as well.

So basically the number should match this regular expression:

[-+]?[0-9]*(.[0-9]+)?(e[-+]?[0-9]+)?

public boolean isNumber(String s) {
    s = s.trim();

    boolean pointSeen = false;
    boolean eSeen = false;
    boolean numberSeen = false;
    boolean numberAfterE = true;
    for(int i=0; i<s.length(); i++) {
        if('0' <= s.charAt(i) && s.charAt(i) <= '9') {
            numberSeen = true;
            numberAfterE = true;
        } else if(s.charAt(i) == '.') {
            if(eSeen || pointSeen) {
                return false;
            }
            pointSeen = true;
        } else if(s.charAt(i) == 'e') {
            if(eSeen || !numberSeen) {
                return false;
            }
            numberAfterE = false;
            eSeen = true;
        } else if(s.charAt(i) == '-' || s.charAt(i) == '+') {
            if(i != 0 && s.charAt(i-1) != 'e') {
                return false;
            }
        } else {
            return false;
        }
    }

    return numberSeen && numberAfterE;
}