給定某一長字符串s='xxxxbobobxxxx',試判斷ss='bob'在s中出現的次數。結果為2
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Jeffacode
TA貢獻1條經驗 獲得超1個贊
def?frequency(s1,?s2): count?=?0 while?s1: if?s1.find(s2)?==?-1: break count?+=?1 s1?=?s1[s1.find(s2)?+?len(s2):] return?count if?__name__?==?'__main__': s1?=?"xxxxbobobxxxx" s2?=?"bob" print("%s?occurs?%d?times?in?%s"?%?(s2,?frequency(s1,?s2),?s1))

小蘿卜腿
TA貢獻51條經驗 獲得超20個贊
#已經解決了 def?finds(s1,?s2): ????i?=?0 ????count?=?0 ????while(i<len(s2)?-?len(s1)?+?1): ????????if(s1?==?s2[i:i+3]): ????????????count+=1 ????????i+=1 ????return?count s1?=?'bob' s2?=?'azcbobobegghakl' print?("Number?of?times?bob?occurs?is:?"+str(finds(s1,?s2)))
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