## Monday, July 6, 2015

### LeetCode OJ - Reverse Integer

Problem:

Solution:

It is no different from reversing a string, except we need to handle overflow carefully. For addition, checking sign suffice, but for multiplication, sometimes overflow happens without changing sign. Since we only multiply by ten, we can use this to help.

if ((x * 10) %10 == 0) then no overflow occurs.

This works because whenever overflow happens. It is wrapped around 2^31 which is not a multiple of 10!

Code:

#include "stdafx.h"

// https://leetcode.com/problems/reverse-integer/

#include "LEET_REVERSE_INTEGER.h"
#include <map>
#include <iostream>
#include <vector>

using namespace std;

namespace _LEET_REVERSE_INTEGER
{
class Solution
{
public:
int reverse(int x)
{
int num_digits = 1;
int y = x;
while (y / 10 != 0)
{
num_digits++;
y = y / 10;
}
try
{
return reverse_helper(x, num_digits);
}
catch (int& x)
{
return 0;
}
}
private:
int shift(int digit, int num_digits)
{
if (num_digits == 1)
{
return digit;
}
else
{
int recursive_result = shift(digit, num_digits - 1);
int product = recursive_result * 10;
if (product % 10 != 0)
{
throw 0;
}
else
{
return product;
}
}
}

{
int sum = x + y;
if (x > 0 && y > 0 && sum < 0)
{
throw 0;
}
else if (x < 0 && y < 0 && sum > 0)
{
throw 0;
}
else
{
return sum;
}
}

int reverse_helper(int num, int num_digits)
{
if (num_digits == 1)
{
return num;
}
else
{
return add(reverse_helper(num / 10, num_digits - 1), shift(num % 10, num_digits));
}
}
};
};

using namespace _LEET_REVERSE_INTEGER;

int LEET_REVERSE_INTEGER()
{
cout << (1 << 31) << endl;
Solution s;
cout << s.reverse(123) << endl;
cout << s.reverse(-123) << endl;
cout << s.reverse(0) << endl;
cout << s.reverse(100) << endl;
cout << s.reverse(101) << endl;
cout << s.reverse(1010) << endl;
cout << s.reverse(1534236469) << endl;
cout << s.reverse(1 << 31) << endl;
cout << s.reverse(~(1 << 31)) << endl;
return 0;
}