首页 > itarticle > [leetcode] problem 65 – valid number

[leetcode] problem 65 – valid number

admin 11月 13, 2020 0

Validate if a given string is numeric.

Example

“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.

Code

public boolean (String s) {
    boolean isNum = false;
    int i = 0;

    while (i < s.length() && Character.isWhitespace(s.charAt(i)))
        i++;

    if (i < s.length() && (s.charAt(i) == '+' || s.charAt(i) == '-'))
        i++;

    while (i < s.length() && Character.isDigit(s.charAt(i))) {
        isNum = true;
        i++;
    }

    if (i < s.length() && s.charAt(i) == '.') {
        i++;

        while (i < s.length() && Character.isDigit(s.charAt(i))) {
            isNum = true;
            i++;
        }
    }

    if (isNum && i < s.length() && s.charAt(i) == 'e') {
        isNum = false;
        i++;

        if (i < s.length() && (s.charAt(i) == '+' || s.charAt(i) == '-'))
            i++;

        while (i < s.length() && Character.isDigit(s.charAt(i))) {
            isNum = true;
            i++;
        }
    }

    while (i < s.length() && Character.isWhitespace(s.charAt(i)))
        i++;

    return isNum && i == s.length();
}