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-->

x = 10

x = int ('10')

x = int ('10.1')

​

​

# 当然第三个语句会报以下这个错误

#ValueError: invalid literal for int() with base 10: '10.1'

# 所以int括号里是不能加小数的

x

# 可变和不可变都是针对变量值而言的

# 比如以下例程

# 可变(针对变量值而言，如果变量值改变而id内存地址不变，则称为可变的量，比如列表，字典等）

lis = [1,2,3]

print(id(lis))#2978432595720

lis[0]=2

print(id(lis))#2978432595720

#当然以上这个id内存地址不是固定不变的，每次生成都会改变

​

# 不可变(值变id号也变化，也就是说值和id内存空间是绑定的，值一旦改变内存空间也变化了)

x = 10

print(id(x))#1896115520

x += 1

print(id(x))#1896115552

x

x1 = float('1.11')

x2 = float('1')

print(type(x2))

​

输出结果为：

1.11

1.0

<class 'float'>

<class 'float'>

​

​

# 可以看到,上述定义中虽然float里面是1，但是类型还是浮点型，而不是整型

xxxxxxxxxx

s = b'ab'#此处留意，b开头在字符串前，打印出来会是bytes类型，也就是二进制类型，不是十进制

print(s)

print(type(s))

​

#其输出结果为：

b'ab'

<class 'bytes'>

​

x

s = 'dog is cut'

# 索引号自左至右，从0开始

# 最常用的内置方法

#1. 索引取值

print(s[1])

o

​

# 2. 切片

print(s[0:4:1])#1表示从左向右切片

`dog `

print(s[4::-1])# -1表示从右向左,不推荐用

​

# 3.for循环

for i in s:

print(i)

​

# 4.striip()# 去两端指定字符，不能去中间的

s1 = '      nick handsome'

print(s1.strip())

print(s1.strip(' niem'))

s2 = '**!!nick handsome'

print(s2.strip('-*!e'))# 首先判断字符串s2的两端字符，为*，然后去strip里面找有没有*，有就去掉，再去判断字符串s2里的-!，以此类推

​

#制定多个字符一起去掉，只要strip里面有的字符就全部干掉

​

#5. split()切割，断开

print(s2.split())#默认以空格切割字符串

print(s2.split('!'))#以制定字符切割

print(s2.split('!',1))#切割一次，数字就是次数

​

#6. in 或not in

print('*'in s2)

​

#7. 长度len

print(len(s2)) #求字符串的长度

lis = ['sfd','sdf','sdfsf','sfdf']

dict = {'a':1,'b':55,'c':464}

print(len(s1))

print(len(dict))

print(len(lis))

​

# 稍微没那么常用的内置方法

# 1. lstrip() 左切割

# 2. rstrip()右切割，用法等同于strip

​

#

# 3. lower&upper全部大写，全部小写

s3 = 'aaabbJ'

print(s3.lower())

print(s3.upper())

# 4. startswith&endswith

# 判断字符串开头和结尾是不是某字符，返回值是True或者False

​

# 5. rsplit右切割

​

# 6. join# 用的较多的，增加字符串

print(s3.join(['353','345','53453','3534']))# 以s3位间隔符，拼接列表里每一个元素

s = '辣条/薯片/汽水/当归'

s1 = s.split('/'))

' '.join(s1)

print(s)# join和split通常组合使用

# 7. replace#顾名思义，替换

s2 = 'nick handsome'

print(f"{s2.replace('handsome','shuai')}")

​

# 8. isdigit判断是否为纯数字，salpha()判断是否为纯字符

s2 = 'aaccc'

print(s2.isdigit())

print(s2.isalpha())

​

# 比如如果要实现账号密码的登录验证：

username_db = 'nick'

pwd_db = '123'

inp_username = input('username:')

inp_pwd = input('pwd:')

​

​

​

# 不太常用的一些内置方法

​

# find|rfind|index|rindex|count

s2 = '****nick % handsome% ****'

print(s2.find('%'))

print(s2.rfind('%')) # 从左找和从右找，找到即停止，只限第一个

print(s2.index('%'))

print(s2.rindex('%'))#index找不到会报错，但是find找不到会返回值-1，所以日后编程多用find，而不是index

​

# center|ljust|rjust|z

s2 = 'nick handsome'

print(s2.center(50,'*'))# 填充到50字符，居中

print(s2.ljust(50,'*'))#居左

print(s2.rjust(50,'*'))#居右

​

# fill

print(s2.zfill(50))#填充0居右

​

# expandtabs

s2 = 'a\ta'# \t是缩进四个字符

print(s2)

print(s2.expandtabs(32))# 针对\t扩充tab

​

# captalize|swapcase|title

s2 = 'harry Potter'

​

print(s2.capitalize())#首字母大写，用在段落开始

print(s2.swapcase())#大小写互换

print(s2.title())#所有单词首字母大写，仅针对英文，对中文无效

​

xxxxxxxxxx

class Solution:

    def romanToInt(self, s: str) -> int:

        s1 = list(s)

        d = 0

        size = len(s)

        s1.append('0')

        count = 0

        for i in range(size):

            if count == i+1 :

                continue

            if s1[i] == 'I':

                if s1[i+1] == 'X':

                    d += 9

                    count += 2

                elif s1[i+1] == 'V':

                    d += 4

                    count += 2

                else:

                    d += 1

                    count += 1

            if s1[i] == 'X':

                if s1[i+1] == 'C':

                    d += 90

                    count += 2

                elif s1[i+1] == 'L':

                    d += 40

                    count += 2

                else:

                    d += 10

                    count += 1

            if s1[i] == 'C':

                if s1[i+1] == 'M':

                    d += 900

                    count += 2

                elif s1[i+1] == 'D':

                    d += 400

                    count += 2

                else:

                    d += 100

                    count += 1

            if s1[i] == 'V':

                d += 5

                count += 1

            if s1[i] == 'L':

                d += 50

                count += 1

            if s1[i] == 'D':

                d += 500

                count += 1

            if s1[i] == 'M':

                d += 1000

                count += 1

        return(d)

xxxxxxxxxx

class Solution:

    def romanToInt(self, s: str) -> int:

                    my_dict = {'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000,'IV':4,'IX':9,'XL':40,'XC':90,'CD':400,'CM':900,'0':0}

                    s1 = list(s)

                    d = 0

                    size = len(s)

                    count = 0

                    s1.append('0')

                    for i in range(size):

                        if count == i + 1:

                             continue

                        if my_dict[s1[i]] < my_dict[s1[i+1]]:

                            d += my_dict[f'{s1[i]}{s1[i+1]}']

                            count+=2

                        else:

                            d += my_dict[s1[i]]

                            count+=1

                    return(d)

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