Thanks Ron,
surely set is the simplest way to understand the question, to see
whether there is a non-empty intersection. But I did the following
thing in a silly way, still not sure whether it is going to be linear
time.
def foo():
l = [...]
s = [...]
dic = {}
for i in l:
dic[i] = 0
k=0
while k <len(s):
if s[k] in dic:
return True
else: pass
k+=1
if k == len(s):
return False
I am still a rookie, and partly just migrated from Haskell...
I am not clear about how making one of the lists a dictionary is
helpful
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