【算法题解】正则表达式操练起来！

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题目

65. Valid Number

A valid number can be split up into these components (in order):

1. A decimal number or an integer.
2. (Optional) An 'e' or 'E', followed by an integer.

A decimal number can be split up into these components (in order):

1. (Optional) A sign character (either '+' or '-').

2. One of the following formats:

   1. One or more digits, followed by a dot '.'.
   2. One or more digits, followed by a dot '.', followed by one or more digits.
   3. A dot '.', followed by one or more digits.

An integer can be split up into these components (in order):

1. (Optional) A sign character (either '+' or '-').
2. One or more digits.

For example, all the following are valid numbers: ["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"], while the following are not valid numbers: ["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"].

Given a string s, return true if s is a valid number.

 

Example 1:

Input: s = "0"
Output: true

Example 2:

Input: s = "e"
Output: false

Example 3:

Input: s = "."
Output: false

 

Constraints:

• 1 <= s.length <= 20
• s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.

题解

题目明确的给出了parse的规则，所以用正则最合适了，直接把题面翻译成正则表达式即可（滑稽 以下为AC代码，附赠一个题目自带的测试。

import re

class Solution:

    def isNumber_test(self, s: str) -> bool:
        ts = ["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"]
        fs = ["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"]
        print([self.isNumber_test(s) for s in ts])
        print([self.isNumber_test(s) for s in fs])

    def isNumber(self, s: str) -> bool:
        re_decimal = "(\+|-)?(([0-9]+\.[0-9]*)|(\.[0-9]+))"
        re_int = "(\+|-)?[0-9]+"
        re_real = "(({0})|({1}))((e|E)({1}))?".format(re_decimal, re_int)
        match_result = re.match(re_real, s)
        # print(match_result)
        try:
            return match_result.span() == (0, len(s))
        except Exception as e:
            return False