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0007._Reverse_Integer.md

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7. Reverse Integer

难度: Easy

刷题内容

原题连接

内容描述

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: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

解题方案

思路1

翻转数字问题需要注意的就是溢出问题，为什么会存在溢出问题呢，我们知道int型的数值范围是 -2147483648～2147483647(负的2的31次方~2的31次方-1)， 那么如果我们要翻转 1000000009 这个在范围内的数得到 9000000001，而翻转后的数就超过了范围。

如果输入的是负数，就递归调用原函数，参数变成-x即可

class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        if x < 0:
            return -self.reverse(-x)
        res = 0
        while x:
            res = res * 10 + x % 10
            x /= 10
        return res if res <= 0x7fffffff else 0

思路2

按照参数正负号先将其转成字符串，然后再反转，根据是否溢出决定输出0还是反转结果

class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """    
        x = -int(str(x)[::-1][:-1]) if x < 0 else int(str(x)[::-1])   # [:-1]相当于把负号去掉
        x = 0 if abs(x) > 0x7FFFFFFF else x
        return x

思路3(StefanPochmann大神)：

看这个解法前先看backticks

cmp函数在python3.x中用不了了，import operator用gt或者lt吧，或者回归if/else condition爸爸的怀抱吧！

class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        s = cmp(x, 0)
        r = int(`s * x`[::-1])
        return s * r * (r < 2 ** 31)

思路4

  • 1.记录符号
  • 2.将数字当字符串处理
  • 3.判断是否阈值区间，进行比较就行
class Solution:
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        mark = 1 if x>=0 else -1
        x_abs = abs(x)
        result = mark * int(str(x_abs)[::-1])
        return result if -2**31 <= result <= 2**31-1 else 0