Leetcode_Reverse Integer（考虑了溢出情况）

Reverse Integer Total Accepted: 52886 Total Submissions: 174213 My Submissions Question Solution

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.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

#include <iostream>
using namespace std;
class Solution {
/*本题的难度在于确定倒置之后，数据是否会溢出。
这里的判断方法是：将x倒置之后得到sum，sum再倒置一次得到r，
判断x和r是否相等，相等，则没有溢出；不相等，可分为两组情况：
1.没有溢出，但是由于x的后面位数是0，导致经过两次倒置，r与x不等；
2.溢出，直接输出0
*/
public:
int reverse(int x) {

int sum=rever(x);
int r=rever(sum);
if(x!=r)
{
while(x%10==0)
{
x=x/10;
}
if(x==r)
return sum;
return 0;
}

return sum;
}
private:
int rever(int x)
{
int sum=0;
int i=x;
while(i!=0)
{
sum=sum*10+i%10;
i=i/10;
}
return sum;
}
};

int main()
{
int x;
cin>>x;
Solution S;
cout<<S.reverse(x);
return 0;
}