LeetCode 7. 整数反转

Posted 2021-12-28 Henny223

7. 整数反转

Code

法一：数学

时间复杂度：O(log|x|)
空间复杂度：O(N)

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

        res = 0
        flag = 0
        if(x < 0): 
            x = -x
            flag = 1 
        length = len(str(x))
        for i in range(length):
            if res < MIN // 10 + 1 or res > MAX // 10: # 注意负数整除要加一
                return 0
            ele = x % 10 # 取最后一位数
            x = x // 10 # 删掉最后一位数
            res = res * 10 + ele # 把删掉的数加进去
        if(flag):
            res = -res

        return res

法二：字符串切片反转

class Solution:
    def reverse(self, x: int) -> int:

        y = int(str(abs(x))[::-1])
        if y < (-2**31) or y > (2**31 - 1):
            return 0
        if x < 0:
            y = -y
        return y