给定某一长字符串s='xxxxbobobxxxx',试判断ss='bob'在s中出现的次数。结果为2
2 回答
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Jeffacode
TA贡献1条经验 获得超1个赞
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))
小萝卜腿
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#已经解决了
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)))添加回答
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