Re: [討論] 用python找出一串數字中最長的"二數數串"
numstring = '889988899278392520771323543282829292222943485709'
results = []
numset = set()
for start in range(len(numstring)):
for end in range(start,len(numstring)):
numset.add(numstring[end])
if len(numset)>2:
break
results.append((start,end))
numset.clear()
max_length = max([end-start for (start,end) in results])
max_intervals = [(start,end) for (start,end) in results if
end-start==max_length]
for start,end in max_intervals:
print numstring[start:end]
※ 引述《uranusjr (←這人是超級笨蛋)》之銘言：
: 我前陣子 (PyConTW 的時候) 才知道 collections 有一個很 IMBA 的東西叫 Counter
: import sys
: from collections import Counter
: numstr = sys.argv[-1] # 這裡讀入要算的字串
: matches = ['']
: for start in range(len(numstr)):
: for end in range(start + 1, len(numstr) + 1):
: substring = numstr[start:end]
: counter = Counter(substring)
: if len(counter) != 2:
: continue
: if len(matches[0]) < len(substring):
: matches = [substring]
: elif len(matches[0]) == len(substring):
: matches.append(substring)
: print('{count} matches (of length {length}):'
: .format(count=len(matches), length=len(matches[0])))
: for match in matches:
: print match
: 淺顯易懂
: ※ 引述《dadadavid (大大大衛)》之銘言：
: : numstring = "889988899278392520771323543282829292222943485709"
: : def ngram(n, iter_tokens):
: : """Return a generator of n-gram from an iterable"""
: : z = len(iter_tokens)
: : return (iter_tokens[i:i+n] for i in xrange(z-n+1))
: : for i in range(len(numstring), 1, -1):
: : results = [ng for ng in ngram(i, numstring) if len(set(ng))==2]
: : if len(results) > 0:
: : print results
: : break
--
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