LeetCode Reverse Integer

 Problem Description: https://leetcode.com/problems/reverse-integer/description/

Accepted Solution

/*

I have learned some crucial things during solving the problem. For example,

if you want to check integer overflows for several operations, you have to manage the following conditions:

• int a = <something>; int x = <something>;
• if (x < 0 && a > INT_MAX + x) // a - x would overflow // `a - x` would overflow
• if (x > 0 && a < INT_MIN + x) // a - x would underflow

• if (a == -1 && x == INT_MIN) // a * x can overflow

• if (x == -1 && a == INT_MIN) // a * x (or a / x) can overflow

• if (x != 0 && a > INT_MAX / x) // a * x would overflow

• if (x != 0 && a < INT_MIN / x) // a * x would underflow

Hence, check the appropriate conditions in perfect positions at your code. Also, keep your attention when you are dealing with a negative number. Moreover, a negative number is unsuitable for direct calculation or dividing by 10. Therefore, when the 'x' is negative but less than INT_MAX, in that case, make it positive first, reverse the digits and return finally it as a non-positive value.

*/

class Solution {

public:

const int high = 1 << 31 - 1;

int power(int p)

{

    int res = 1;

    for(int i=0; i<p; i++)

    {

        res *= 10;

    }

    //cout << "res = " << res << "\n";

    return res;

}

int reverse(int x) {

    int ar[15], len=0, p, q;

    //cout << "Enter x: ";

    //cin >> x;

    //x = (1 << 31)-1;

    //cout << "main = " << x << "\n";

    bool negative = false;

    bool overflow = false;

    if(x < 0)

    {

        negative = true;

        p = -1;

        q = x;

        if(p == -1 && q == INT_MIN) // if negative a*x is overflow

        {

            overflow = true;

        }

        else

        {

            x = x * (-1);

        }

        //cout << x << " " << overflow << "\n";

    }

    while(x > 0)

    {

        ar[len++] = x % 10;

        x /= 10;

    }

    //for(int i=0; i<len; i++) cout << ar[i] << " ";

    int num = 0;

    for(int i=0, j=len-1; i<len; i++, j--)

    {

        p = ar[i];

        q = power(j);

        if(p != 0 && q > INT_MAX / p) // if p * q is overflow

        {

            overflow = true;

            break;

        }

       

        int t = p * q;

        if(num > 0 && t > INT_MAX - num) // if a + x is overflow

        {

            overflow = true;

            break;

        }

        num = num + t;

    }

    if(overflow)

    {

        return 0;

        //cout << 0 << "\n";

    }

    else

    {

        if(negative)

        {

            //cout << "-" << num << "\n";

            num = num * (-1);

            //cout << num << "\n";

            return num;

        }

        else

        {

            //cout << num << "\n";

            return num;

        }

    }

}

};