LeetCode //C - 7. Reverse Integer

7. Reverse Integer

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [ − 2 31 , 2 31 − 1 ] [-2^{31}, 2^{31} - 1] [−231,231−1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Example 1:

Input: x = 123
Output: 321

Example 2:

Input: x = -123
Output: -321

Example 3:

Input: x = 120
Output: 21

Constraints:

• − 2 31 < = x < = 2 31 − 1 -2^{31} <= x <= 2^{31} - 1 −231<=x<=231−1

From: LeetCode

Link: 7. Reverse Integer

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Solution:

Ideas:

1. Initialize a result variable (reversed) to zero: This will hold our reversed number.

2. Loop until x is zero:

• Extract the last digit of x using x % 10.
• Divide x by 10 to remove the last digit.

3. Overflow/Underflow check:

• Before appending a digit to reversed, check if appending it would cause the number to overflow or underflow the 32-bit integer limits (INT_MAX and INT_MIN from limits.h).
• If overflow or underflow is detected, return 0.

4. Construct the reversed number:

Multiply the current reversed by 10 (shift digits left) and add the extracted digit.

Code:

int reverse(int x) {
    int reversed = 0;

    while (x != 0) {
        int digit = x % 10;  // Get the last digit of x
        x /= 10;             // Remove the last digit from x

        // Check for potential overflow/underflow before actually adding the digit
        if (reversed > INT_MAX / 10 || (reversed == INT_MAX / 10 && digit > 7)) {
            return 0;  // Overflow condition for positive numbers
        }
        if (reversed < INT_MIN / 10 || (reversed == INT_MIN / 10 && digit < -8)) {
            return 0;  // Underflow condition for negative numbers
        }

        reversed = reversed * 10 + digit;  // Append the digit
    }

    return reversed;
}