# Leetcode 7. Reverse Integer

## 題解

### 數學

#### 利用 Array 儲存

class Solution:
    def reverse(self, x: int) -> int:
        # Time complexity: O(n)
        # Space complexity: O(n)
        nums = []
        isn = -10 if x < 0 else 10
        while x != 0:
            nums.append(x % isn)
            x //= isn
            if isn == -10:
                x = -x
        output = 0
        while len(nums) > 0:
            num = nums.pop(0)
            n = len(nums)
            output += num * (10 ** n)
        return output if (-2 ** 31) <= output <= (2 ** 31) -1 else 0 # 記得最後要判斷是否有在範圍內

#### 取 log10

import math

class Solution:
    def reverse(self, x: int) -> int:
        # Time complexity: O(n)
        # Space complexity: O(1)
        if x == 0:
            return x
        original_x = x
        x = x if x >= 0 else x * -1
        digits = int(math.log10(x)) + 1
        ans = 0
        while digits:
            ans += ((x % 10) * (10 ** (digits - 1)))
            x //= 10
            digits -= 1
        ans *= -1 if original_x < 0 else 1
        return ans if -2 ** 31 <= ans <= (2 ** 31) - 1 else 0