Solving the 'Reverse Integer' on LeetCode with C++

Problem

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

Answer

class Solution {
public:
    int reverse(int x) {
        long long result = 0;
        while(x) {
            result = result * 10 + x % 10;
            if (result > INT_MAX || result < INT_MIN) {
                return 0;
            }
            x /= 10;
        }
        return result;
    }
};

Let's examine the solution to the problem.
For an example, consider the following:

Input: x = -123
Output: -321

We solve the problem by taking the last digit in order.

Input: x = -123
1: -3
2: -32
3: -321

Output: -321

Now, let's dissect the code:

    int reverse(int x) {
        long long result = 0;

The long long result = 0 is used to store the answer. It might exceed the int range because we're reversing x. Therefore, we use long long.

        while(x) {
            result = result * 10 + x % 10;

As mentioned before, we solve the problem by taking the last digit in order.

We use x % 10 to get the last digit. The variable result will have a value starting from the second iteration. We use result * 10 to shift the digits one place to the left as we add the new digit to the rightmost place.

x = -123
result = 0
Step 1: result = result * 10 + x % 10 = 0 * 10 + -123 % 10 = -3
x = -12 // explained later in this blog post.
result = -3
Step 2: result = result * 10 + x % 10 = - 3 * 10 + -12 % 10 = -32
x = -1
result = -32
Step 3: result = result * 10 + x % 10 = - 32 * 10 + -1 % 10 = -321

            if (result > INT_MAX || result < INT_MIN) {
                return 0;
            }

If result exceeds the range of int, we must return 0. The condition result > INT_MAX || result < INT_MIN checks for that.

            x /= 10;

We use x /= 10 to eliminate the last digit.

For instance,

x = - 123

Step 1: 
x = -123
// result = 0
result = result * 10 + x % 10 = -3

Step 2: 
x = -12
// result = -3 
result = result * 10 + x % 10 = -3

We need to eliminate the rightmost digit, so we use x /= 10 to shift the digits one place to the right.

We continue this process until x becomes 0.