题目链接

Python3

官方解法 $⟮O(|x|)、O(1)⟯$

class Solution:
    def reverse(self, x: int) -> int:
        INT_MIN, INT_MAX = -2**31, 2**31 - 1

        rev = 0
        while x != 0:
            if rev < INT_MIN // 10 + 1 or rev > INT_MAX // 10:
                return 0 
            
            digit = x % 10 
            if x < 0 and digit > 0:
                digit -= 10  # Python3 的取模运算在 x 为负数时也会返回 [0, 9) 以内的结果，因此这里需要进行特殊判断

            x = (x - digit) // 10
            rev = rev * 10 + digit 

        return rev 

写法2

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

class Solution:
    def reverse(self, x: int) -> int:
        flag = 1
        if x < 0:
            flag = -1
            x = - x
        s = str(x)
        res = ''
        for i in range(len(s)-1, -1, -1):
            res += s[i]

        res = int(''.join(res))
        return 0 if (res > 2**31 - 1 or -res < -2**31) else res * flag     

写法3

class Solution:
    def reverse(self, x: int) -> int:
        flag = 1
        if x < 0:
            flag = -1
            x = - x

        res = 0
        while x > 0:
            x, tmp = divmod(x, 10)
            res = res * 10 + tmp 

        res *= flag 
        if -2**31 <= res <= 2**31 - 1:
            return res
        else:
            return 0

C++

官方解法 $⟮O(|x|)、O(1)⟯$

class Solution {
public:
    int reverse(int x) {
        int res = 0;
        //int INT_MAX = 2**31-1; INT_MIN = -2**31; // C++ 自带
        while (x != 0){
            if (res < INT_MIN / 10 || res > INT_MAX / 10){// 注意写法
                return 0;
            }

            int digit = x % 10;
            x /= 10;
            res = res * 10 + digit;
        }
        return res;
    }
};