Leetcode_num4_Reverse Integer

题目：

Reverse digits of an integer.

Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

每次一遇到关于整型数的题就有些发怵，看来日后还得加强一下~

此题的难点主要在几个可能出现的bug上，如数字末尾带0的要去0再反转，反转后数值溢出的问题（int类型32位，数值范围在-2^31~2^31之间）

思路分为：

1 对数字去除末尾的零

2 将数字转化为字符串处理，由于字符串的不可变性，采用字符串数组存储反转字符串结果

3 通过比较原数字字符串与最大数值‘2147483647’每一位上数字的大小来解决溢出问题

4 将字符串数组中的数字字符相加，转化成为int类型输出

代码如下：

class Solution:
# @return an integer
def reverse(self, x):
if x==0:
return 0
a=0 #dealing with last 0s
b=0
while(b==0):
a=x//10
b=x%10
x=a
x=a*10+b

c=str(x)#convert int into string
L=len(c)
s=['' for i in range(L)]

MAX='2147483647'#handling overflow case
if c.startswith('-'):
l=len(c)
if l>11:
return 0
elif l==11:
for i in range(1,11):
if c[L-i-1]>MAX[i]:
return 0
else:
l=len(s)
if l>10:
return 0
elif l==10:
for i in range(0,10):
if c[L-i-1]>MAX[i]:
return 0

if c.startswith('-'):
s[0]='-'
for i in range(1,L/2+1):
t=c[i]
s[i]=c[L-i]#it's very crucial
s[L-i]=t
else:
for i in range(0,L/2+1):
t=c[i]
s[i]=c[L-i-1]
s[L-i-1]=t
rs=''
for i in s:
rs+=i
return int(rs)