7. 整数反转
方法一：
class Solution:
def reverse(self, x: int) -> int:
flag = 1 if x >= 0 else -1
d = int(str(abs(x))[::-1]) * flag
return d if 2**31 - 1 >= d >= -2**31 else 0方法二：
class Solution:
def reverse(self, x: int) -> int:
res = 0
flag = -1 if x < 0 else 1
y = abs(x)
while y:
y, tem = divmod(y,10)
res = res * 10 + tem
if 2**31 - 1 < res or res * flag < -2**31:
return 0
return res * flag8. 字符串转换整数 (atoi)
方法一：
class Solution:
def myAtoi(self, s: str) -> int:
s = s.strip()
if s == '': return 0
x, tem, flag = 0, '', -1 if s[0] == '-' else 1
# 去掉符号位
# if s[0] == '-' or s[0] == '+': s = s[1:]
if s[0] in ['-','+']: s = s[1:]
for c in s:
if not c.isdigit(): break
tem += c
if tem:
x = int(tem) * flag
if x < -2**31: x = -2**31
if x > 2**31 - 1: x = 2**31 - 1
return x方法二： re
class Solution:
def myAtoi(self, s: str) -> int:
# *args 和 **kwargs 不定长参数，re.findall 结果是列表作为解包提供给 int
return max(min(int(*re.findall('^[\+\-]?\d+', s.lstrip())), 2**31 - 1), -2**31)方法三：
class Solution:
def myAtoi(self, s: str) -> int:
s = s.strip()
if not s : return 0
x, tem = 0, ''
if s[0] in ['-','+']:
tem = s[0]
s = s[1:]
# 去除只有一个符号 '+' 或连续 '+-' 开头的串，如果第一个是符号，则第二个必须是数字。
if not s or not s[0].isdigit(): return 0
for c in s:
if not c.isdigit(): break
tem += c
return max(min(int(tem),2**31 - 1),-2**31)
