Leetcode: Valid Number

最新推荐文章于 2025-08-08 21:26:37 发布

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本文介绍了一种判断给定字符串是否表示数值的方法，包括整数、小数和科学计数法表示的数值。通过解析字符串中的字符，确保其符合特定的格式要求。

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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.

1. all the characters must be numbers, dot, 'e', +/- and white space

2. leading and ending white spaces are allowed

3. string could start with +/-

4. if +/- is in the middle, it must has a 'e' right before where +/- is

5. there shouldn't be more than one 'e' and one dot

6. 'e' must follow a number

7. dot must go before 'e'

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