【LeetCode】7. 整数反转

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 - 1rev = 0while 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) // 10rev = rev * 10 + digit return rev 

写法2

class Solution:def reverse(self, x: int) -> int:flag = 1if x < 0:flag = -1x = - xres = 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 = 1if x < 0:flag = -1x = - xs = 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 = 1if x < 0:flag = -1x = - xres = 0while x > 0:x, tmp = divmod(x, 10)res = res * 10 + tmp res *= flag if -2**31 <= res <= 2**31 - 1:return reselse: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;}
};