leetcode 7 Reverse Integer

 

7

Reverse Integer

24.40%

反转数字，注意首位为0

32bit：数字大小小于2^31

string反转方法:[::-1]

 

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Sun Dec 24 20:13:51 2017

@author: vicky
"""

#102ms
#import numpy as np;
import math

#self 代表类的实例，self 在定义类的方法时是必须有的，虽然在调用时不必传入相应的参数。
#没有class类，只有def函数时，不用self
class Solution:
    def reverse(self, x): 
        """
        :type x: int
        :rtype: int
        """
        s2=0
        if x==0:
            s2=0
        else:
            k=int(math.log10(abs(x)))
            s=abs(x)
            x2={} #提取各位数字
            for i in range(k+1):
                x2[i]=int(s/(10**(k-i)))
                s=s-x2[i]*(10**(k-i))
            for i in range(0,k+1):
                s2=s2+x2[k-i]*(10**(k-i))
            if x<0:
                s2=-1*s2
        return s2 if abs(s2)<2**31 else 0 #32-bit意思是小于数字大小小于2^32
        
x=-15342360
t=Solution()
t.reverse(x)
print(t.reverse(x))

#import operator
#def reverse( x):
#    s = operator.lt(x, 0)
#    r = int('s*x'[::-1])
#    return s*r * (r < 2**31)
#        
#reverse(1534236469)

#98ms
class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        #先把转换为str型，再转换为int型，不用考虑首位为0因为int
        x = int(str(x)[::-1]) if x >= 0 else -int(str(-x)[::-1])
        return x if x < 2147483648 and x >= -2147483648 else 0
    
#99ms
class Solution:
    def reverse(self, x): 
        """
        :type x: int
        :rtype: int
        """
        s3=abs(x)
        sign_x=1 if x>0 else -1
        new_x=0
        while s3>0:
           new_x=10*new_x+s3%10 
           s3=int(s3/10)
        return sign_x*new_x if new_x<2**31 else 0