题目描述：
Reverse digits of an integer.
Example1:
x = 123, return 321
Example2:
x = -123, return -321
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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.
Note: The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
思路：
本题重点关注溢出问题，可通过 long long 来避免运算过程中导致溢出，最终判断结果时通过 INT_MAX 或 INT_MIN 判断溢出。
代码实现：
class Solution {
public:
int reverse(int x) {
int minus = 0;
if(x<0){
minus = 1;
x = 0-x;
}
long long t=0;
while(x!=0){
t = x%10+10*t;
x/=10;
}
t = minus==1?(0-t):t;
return (t>INT_MAX||t<INT_MIN)?0:t;
}
};