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[LeetCode] 65. Valid Number

SoapEye / 2786人阅读

Problem

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

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.

Solution
class Solution {
    public boolean isNumber(String s) {
        if (s == null || s.length() == 0) return false;
        
        s = s.trim();
        int len = s.length();
        if (len == 0) return false;
        
        int signCount = 0;
        boolean hasE = false;
        boolean hasNum = false;
        boolean hasPoint = false;
        
        for (int i = 0; i < len; i++) {
            char ch = s.charAt(i);
            
            if (ch >= "0" && ch <= "9") {
                hasNum = true;
            } else if (ch == "e" || ch == "E") {
                if (hasE || !hasNum || i == len-1) return false;
                hasE = true;
            } else if (ch == ".") {
                if (hasPoint || hasE) return false;
                //0. is true
                if (i == len-1 && !hasNum) return false;
                hasPoint = true;
            } else if (ch == "+" || ch == "-") {
                //two signs at most; should not appear in the end
                if (signCount == 2 || i == len-1) return false;
                //sign appears in the middle but no E/e
                if (i > 0 && !hasE) return false;
                signCount++;
            } else return false;
        }
        
        return true;
    }
}

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