[leetcode]65. Valid Number-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_36869329/article/details/85221004

本文深入探讨了如何通过正则表达式验证并解析浮点数字符串，包括整数部分、小数部分及科学计数法的处理。文章详细解释了每个字符在浮点数中的作用，如数字、小数点、正负号和指数符号，并提供了相应的Java代码实现。

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We start with trimming.

If we see [0-9] we reset the number flags.
We can only see . if we didn't see e or ..
We can only see e if we didn't see e but we did see a number. We reset numberAfterE flag.
We can only see + and - in the beginning and after an e
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]*)?)|.[0-9]+)(e[-+]?[0-9]+)?

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

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