LeetCode--Reverse Integer(整数反转)Python
2017-12-05 15:41
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题目:
给定一个整数,返回反转后的整数。如给定12,返回21,给定210返回12,给定-231返回-132.
解题思路:
将整数先转为字符串格式。再倒序对字符串进行读取。
代码(Python):
class Solution(object):
def reverse(self, x):
"""
:type x: int
:rtype: int
"""
# if x<-2^31 or x>2^31-1:
# return 0
count = 0
s = str(x)
if s[0]=='-':
for i in range(len(s)-1):
count = count*10+int(s[-1-i])
if count>2**31:
return 0
return -count
else:
for i in range(len(s)):
count = count*10+int(s[-i-1])
if count>2**31-1:
return 0
return count
给定一个整数,返回反转后的整数。如给定12,返回21,给定210返回12,给定-231返回-132.
解题思路:
将整数先转为字符串格式。再倒序对字符串进行读取。
代码(Python):
class Solution(object):
def reverse(self, x):
"""
:type x: int
:rtype: int
"""
# if x<-2^31 or x>2^31-1:
# return 0
count = 0
s = str(x)
if s[0]=='-':
for i in range(len(s)-1):
count = count*10+int(s[-1-i])
if count>2**31:
return 0
return -count
else:
for i in range(len(s)):
count = count*10+int(s[-i-1])
if count>2**31-1:
return 0
return count
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