>>> for subLi in x:
... if(xdictstart.has_key(subLi[0])):
... if(xdictstart[subLi[0]]>subLi[1]):
...xdictstart[subLi[0]]=subLi[1]
... else:
... xdictstart[subLi[0]]=subLi[1]
The opposite can be done to obtain the xdictend.This is one way.There might
be some easier way's but i'm not aware
Below given is solution to a puzzle(
http://projecteuler.net/index.php?section=problems&id=14) in python and c
Python:
import time
startT=time.time()
maxlen=0
longest=0
for i in xrange(1,100):
last=i
cnt=0
while(last <> 1):
cnt=cnt+1
if(last%2==0):
last=last/2
else:
last=3*last
This implementation is really good.It's really fast compared to the initial
one I posted but i didn't understand much about this memoize.I asked one of
my friend he told it's python decorators.Can anyone please explain what the
function memoize does.
>
> > Readability counts. Here is my attempt.
