python实现leetcode之65. 有效数字

解题思路

代码比较长，但是很简单
「设计模式」中的访问者模式
使用一个类管理匹配，检查匹配状态即可

65. 有效数字

代码

class Solution:
    def isNumber(self, s: str) -> bool:
        mn = MatchNumber(s)
        mn.match()
        return mn.is_match()


class MatchNumber:
    def __init__(self, s):
        self.s = s
        self.index = 0
        self.fail = False

    def match(self):
        self.match_number()
        x = self.match_exp()
        if x:
            self.match_int()

    def is_match(self):
        return not self.fail and self.index == len(self.s)

    def match_number(self):
        self.match_sign(False)
        x = self.match_digits(False)
        self.match_point(False)
        self.match_digits(not x)

    def match_int(self):
        self.match_sign(False)
        self.match_digits(True)

    def match_sign(self, force=False):
        if self.index < len(self.s) and self.s[self.index] in '-+':
            self.index += 1
            return True
        elif force:
            self.fail = True

    def match_point(self, force=False):
        if self.index < len(self.s) and self.s[self.index] == '.':
            self.index += 1
            return True
        elif force:
            self.fail = True

    def match_digits(self, force=False):
        idx = self.index
        if force and self.index >= len(self.s):
            self.fail = True
        elif force and not self.s[self.index].isdigit():
            self.fail = True
        else:
            while self.index < len(self.s) and self.s[self.index].isdigit():
                self.index += 1
        return idx != self.index

    def match_exp(self, force=False):
        if self.index < len(self.s) and self.s[self.index] in 'eE':
            self.index += 1
            return True
        elif force:
            self.fail = True