reverse_integer.cpp

/*
 *leetcode题

Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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).

 */

#include <iostream>

using namespace std;

class Solution {
    public:
        int reverse (int x) {
            long long r = 0;
            long long t = x;
            t = t > 0 ? t : -t;
            for (; t; t /= 10)  //此处设计精妙，不必使用额外的空间如栈保存每一位上的数
                r = r * 10 + t % 10;
            bool sign = x > 0 ? false: true;
            if (r > 2147483647 || (sign && r > 2147483648)) {
                return 0;
            } else {
                if (sign) {
                    return -r;
                } else {
                    return r;
                } 
            }
        } 
};

int main()
{
    class Solution t;

    cout<<"result "<<t.reverse(1000)<<endl;
}