LeetCode in Kotlin: 7. Reverse Integer

Problem

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

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321
Example 2:

Input: -123
Output: -321
Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Solution

fun reverse(x: Int): Int {
    val str = x.toString()
    val stack = Stack<Char>()
    var result = ""

    str.forEach {
        stack.push(it)
    }

    while (true) {
        if (stack.size == 0) break

        if (result.isEmpty() && stack.peek() == '0') {
            if (stack.size == 1) {
                result = "0"
                break
            }

            stack.pop()
            continue
        }

        if (stack.peek() == '-') {
            result = stack.peek() + result
            break
        }

        result += stack.pop()
    }

    return try {
        result.toInt()
    } catch (e: NumberFormatException) {
        0
    }
}

fun reverse(x: Int): Int {
    val output = StringBuilder()
    var currVal = x

    if (currVal < 0) {
        currVal *= -1
        output.append("-")
    }

    output.append(currVal.toString().reversed())

    return output.toString().toIntOrNull() ?: 0
}

fun reverse(x: Int): Int {
    val y = abs(x)
        .toString()
        .reversed()
        .toIntOrNull() ?: return 0

    return when {
        0 <= x -> y
        else -> y * -1
    }
}

Test code

@Test
fun reverse() {
    assertEquals(321, solution.reverse(123))
    assertEquals(-321, solution.reverse(-123))
    assertEquals(21, solution.reverse(120))
    assertEquals(21, solution.reverse(1200))
    assertEquals(0, solution.reverse(0))
    assertEquals(0, solution.reverse(1534236469))
}

Model solution

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

Approach 1: Pop and Push Digits & Check before Overflow

class Solution {
    public int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int pop = x % 10;
            x /= 10;
            if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
            if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
            rev = rev * 10 + pop;
        }
        return rev;
    }
}