Python 练习题之一回文数

题目：求大于10的，10进制，8进制，2进制表示的回文数都一样的最小的整数。（题目来源程序员的算法趣题）

回文数：将某个数字的每个位的数字反向排列形成的新数和原来的数一样的数，例如9，例如12355321等。

直接出代码：

import numpy as np
import datetime

def calintnum_10(num):
    allnum=[]   
    upvalue=num%10
    allnum.append(upvalue)
    num=int(num/10)
    while(num!=0):
        allnum.append(num%10)      
        num=int(num/10)        
    return allnum  
def calintnum_8(num):
    allnum=[]   
    upvalue=(num)%8
    allnum.append(upvalue)
    num=eval(oct(int((num)/8)))
    while(num!=0):
        allnum.append((num)%8)      
        num=eval(oct(int((num)/8)))    
    return allnum    
def calintnum_2(num):
    allnum=[]   
    upvalue=num%2
    allnum.append(upvalue)
    num=eval(oct(int((num)/2)))
    while(num!=0):
        allnum.append((num)%2)      
        num=eval(oct(int((num)/2)))    
    return allnum 
def judgesuitnum(allnum_ori):    
    allnum_rev=allnum_ori.copy()
    allnum_rev.reverse()
    if(allnum_ori==allnum_rev):
        return True
    else:
        return False
 
    
    
def judgesuitnum3(num):
    num_8=oct(num)
    num_2=bin(num)
    all_10=calintnum_10(num)
    all_8=calintnum_8(eval(num_8))
    all_2=calintnum_2(eval(num_2))   
    
    if(judgesuitnum(all_10) and judgesuitnum(all_8) and judgesuitnum(all_2)):
        return True
    else:
        return False
start = datetime.datetime.now()    
find=False  
num=10          
while(not find):
      find=judgesuitnum3(num)   
      if(find):
          break
      else:          
        num=num+1
      
end=datetime.datetime.now()      
spendtime=end-start
print("find num is:%d"%num)         
print("find Spend time:{}".format(spendtime))     


def strcode(num,cut=True):
    numstr=str(num)
    if(cut):
        numstr=numstr[2:]
    numstr_inv=numstr[::-1] 
    return [numstr,numstr_inv]      

def judgesuitnum3_2(num):
    num_10str=strcode(num,False)
    num_8=oct(num)
    num_8str=strcode(num_8)
    num_2=bin(num)
    num_2str=strcode(num_2)
    
    if(num_10str[0]==num_10str[1] and num_8str[0]==num_8str[1] and num_2str[0]==num_2str[1]):
        return True
    else:
        return False
print("******************************************")      
start = datetime.datetime.now()    
find=False  
num=10          
while(not find):
      find=judgesuitnum3_2(num)   
      if(find):
          break
      else:          
        num=num+1
      
end=datetime.datetime.now()      
spendtime=end-start
print("find num is:%d"%num)         
print("find Spend time:{}".format(spendtime)) 

一共两个方法，写的low,新学的。结果如下：明显第二种方法快了好多。

 欢迎交流。