LeetCode Reverse Integer

大家好，我是刘天昊，这次的题目是比较简单的整数逆序

先看题目

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321
题目还有些提示
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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
说下首先要注意的是末尾是0怎么去翻转
然后要考虑溢出的问题--这个先思考一下（如何去判断一个整数翻转后溢出）

还有处理负数的时候要考虑符号的问题

int reverse(int x)
{
int res=0,rev=0,wei=0,cpres=0;
int cpx=x;
int danger=0;
while(x)
{
res=res*10+x%10;
x/=10;
wei++;
}
cpres=res;
while(cpres)
{
rev=rev*10+cpres%10;
cpres/=10;
}
if(rev!=cpx&&wei>=9)
{
return 0;
}
return res;
}

说下最主要的是如何判断一个数是否溢出，主要是先翻转一次，再翻转一次看看与原数是否相等。相等没有溢出，不想等溢出