Longest Collatz sequence

问题：

The following iterative sequence is defined for the set of positive integers:

n

n/2
(n is even)

n

3n +
1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:

13

40

20

10

5

16

8

4

2

1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at
1.
Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

代码：

import urllib
###############主程序###################
known={1:1}
findnext={2:1}
def Getnext (n):
if n in findnext:
return findnext

else:
nextnum=0
if n%2:
nextnum=3*n+1
else:
nextnum=n/2
findnext
=nextnum
return nextnum

def maxfact (x,y):
if x[0]>y[0]:
return x
else:
return y

def fact (n):
if n in known:
return known

else:
temp=0
nextnum=0
"""
con_action=[(lambda x: not x%2,lambda x:x/2),
(lambda x: x%2,lambda x:3*x+1)]
for indice,(con,action) in enumerate(con_action) :
if con(n):
nextnum= action(n)
break
"""
nextnum=0
if n%2:
nextnum=3*n+1
else:
nextnum=n/2
temp+=1
temp+=fact(nextnum)
known
=temp
return temp
if __name__=="__main__":
li=(1,fact(1))
"""
lishow=[0]*2000000
lindex=0
endindex=int(1000000-1)/3
for rindex in xrange(1000000,endindex,-1):
if not     lishow[rindex]:
thisvalue=fact(rindex)
if thisvalue>li[1]:
li=(rindex,thisvalue)
s_index=rindex
while s_index !=1:
tempvalue=lishow[s_index]
if tempvalue==1:
break
lishow[s_index]=1
s_=Getnext(lindex)
print li
"""
li=reduce(lambda x,y:x if x[0]>y[0]  else y , [(fact(x),x) for x in xrange(1,1000000) ])
#li.sort(reverse=True)
#print known[known.keys()[-2]]
print li

时间是4S-5S，好慢，艹了。。。

大牛的代码：

。。。。好像跟我的都不差不多，艹了。。。