0007 - Reverse Integer (Medium)

Problem Link

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

Problem Statement

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], 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

Approach 1: Standard

As input can be from Integer.MIN_VALUE to Integer.MAX_VALUE.

Reversing some of the 10 digits number, will not provide the correct answer, because the integer range might overflow, if it does return 0, otherwise the reversed number.

For example, reversing the last integer number 2147483647 can go out of range, so having long primitive in the program can hold the range.

First, get the sign bit value, and negate the input if it's negative. Reverse the integer and if it's above max range then return 0 or return the value with sign bit value.

Java

Written by @vigneshshiv

// Time complexity: O(1), where # digits are always <= 10
// Space complexity: O(1)
class Solution {
    public int reverse(int x) {
        long num = 0, remainder = 0;
        int sign = 1;
        if (x < 0) {
            x *= -1;
            sign = -1;
        }
        while (x > 0) {
            remainder = x % 10;
            num = (num * 10) + remainder;
            x /= 10;
        }
        return num > Integer.MAX_VALUE ? 0 : (int) num * sign;
    }
}

JavaScript

Written by @MithunPrabhu777

/**
 * @param {number} x
 * @return {number}
 */
var reverse = function(x) {
    let [number, sign, remainder] = [0, 1, 0];
    const RANGE = 2 ** 31 - 1;
    if (x < 0) {
        x *= -1;
        sign = -1;
    }
    while (x > 0) {
        remainder = x % 10;
        number = (number * 10) + remainder;
        x = Math.floor(x / 10);
    }
    return number > RANGE ? 0 : number * sign;
};

Python

Written by @radojicic23

class Solution:
    def reverse(self, x: int) -> int:
        number, remainder, sign = 0, 0, 1
        MAX = 2 ** 31 - 1
        if x < 0:
            x *= -1
            sign = -1
        while x > 0:
            remainder = x % 10
            number = (number * 10) + remainder
            x //= 10
        return 0 if number > MAX else int(number) * sign

C++

Written by @radojicic23

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