題目鏈接
tag:

• easy

question:
??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: -2147483648～2147483647. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

C++ 解法：
思路：
??用long long型變量保存計算結(jié)果，最后返回的時候判斷是否在int返回內(nèi)，代碼如下：

class Solution {
public:
    int reverse(int x) {
        // 數(shù)學(xué)分析不等式條件
        int res = 0;
        while (x != 0) {
            if (res > INT_MAX / 10 || res < INT_MIN / 10) {
                return 0;
            }
            int digit = x % 10;
            x /= 10;
            res = res*10 + digit;
        }
        return res;
    }
};

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

Python 解法：
??比較簡單，轉(zhuǎn)化為字符串反轉(zhuǎn)即可，見代碼：

class Solution:
    def reverse(self, x: int) -> int:
        res = int(str(abs(x))[::-1])
        if res > 2**31-1:
            return 0
        if x < 0:
            return -res
        else:
            return res