iOS Interview - Leetcode 7: Reverse Integer

April 26, 2024 • 1 min read • #swift, #interview, #leetcode

Primary idea: Using % 10 to reversely iterate through digits of the number, and use * 10 to update the result accordingly
Time Complexity: O(n)
Space Complexity: O(1)

func reverse(_ x: Int) -> Int {
    // make x to be mutable
    var x = x

    // inital value for the result
    var reverseX = 0

    // Iterate until x = 0
    while x != 0 {
        reverseX = reverseX * 10 + x % 10
        x = x / 10

        if reverseX > Int(Int32.max) || reverseX < Int(Int32.min) {
            return 0
        }
    }

    return reverseX
}

print(reverse(123)) // 321
print(reverse(-123)) // -321
print(reverse(120)) // 21
print(reverse(1300)) // 31
print(reverse(1534236469)) // 0
print(reverse(-9223372036854774999)) // 0
print(reverse(9223372036854774999)) // 0
print(reverse(-2147483412)) // -2143847412