题目描述

blablabla

样例

blablabla

算法1

() $O(n^2)$

blablabla

时间复杂度分析：blablabla

Python 代码

class Solution(object):
    def isNumber(self, s):
        """
        :type s: str
        :rtype: bool
        """
        i = 0
        j = len(s) - 1
        while i < len(s) and s[i] == ' ':
            i += 1
        while j > 0 and s[j] == ' ':
            j -= 1
        if i > j:
            return False

        s = s[i:j + 1] # 关键

        if s[0] == '-' or s[0] == '+':
            s = s[1:]
        if len(s) == 0 or s[0] == '.' and len(s) == 1:
            return False

        e = 0
        dot = 0
        i = 0

        while i < len(s): # 此处不可以用for loop 因为i+=1不会改变
            if s[i] >= '0' and s[i] <= '9':
                i += 1
                continue
            elif s[i] == '.':
                dot += 1
                if e or dot > 1: # 如果e后面出现'.' 或者多个'.' False
                    return False
                i += 1
            elif s[i] == 'e' or s[i] == 'E':
                e += 1
                if i + 1 == len(s) or i == 0 or e > 1 or i == 1 and s[0] == '.': # 如果e后面没有数字或.e False
                    return False
                if s[i + 1] == '+' or s[i + 1] == '-': #如果e后面出现符号之后没有数字 False
                    if i + 2 == len(s):
                        return False
                    i += 1
                i += 1
            else:
                return False

        return True