Medium
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 [-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
Example 4:
Input: x = 0
Output: 0
Constraints:
-231 <= x <= 231 - 1
To solve the Reverse Integer problem in Swift using a Solution
class, we’ll follow these steps:
Solution
class with a method named reverse
.rev
), the sign of the input integer (sign
), and the absolute value of the input integer (x
).x
:
rev
by multiplying it by 10 and adding the extracted digit.x
by removing the least significant digit using integer division.rev
overflows the signed 32-bit integer range. If so, return 0.rev
with the appropriate sign.Here’s the implementation:
class Solution {
func reverse(_ x: Int) -> Int {
var rev: Int = 0
var number = x
while number != 0 {
let pop = number % 10
number /= 10
if rev > Int32.max / 10 || (rev == Int32.max / 10 && pop > 7) {
return 0
}
if rev < Int32.min / 10 || (rev == Int32.min / 10 && pop < -8) {
return 0
}
rev = rev * 10 + pop
}
return rev
}
}
This implementation provides a solution to the Reverse Integer problem in Swift.