Python [习题] 求最长共同子串

s1 = 'abcdefg'
s2 = 'defabcdoabcdeftw'
s3 = '1234a'
s4 = 'wqweshjkb'
s5 = 'defabcd'
s6 = 'j'

 

求 s1、s3、s4、s5、s6 分别与 s2 的最长共同子串，分别测试有共同子串和没有共同子串的情况下此函数效率问题。

s1 = 'abcdefg'
s2 = 'defabcdoabcdeftw'
s3 = '1234a'
s4 = 'wqweshjkb'
s5 = 'defabcd'
s6 = 'j'

def findstr(str1,str2):
    '''
    Returns str1 and str2 longest common substring.  2017/10/21 23:30

    :param str1: 'abcdefg'
    :param str2: 'defabcd'
    :return: 'abcd'
    '''
    count = 0
    length = len(str1)
    for sublen in range(length,0,-1):
        for start in range(0,length - sublen + 1):
            count += 1
            substr = str1[start:start+sublen]
            if str2.find(substr) > -1:
                print('count={}  subStringLen:{}'.format(count,sublen))
                return substr
    else:
        return "'{}' and '{}' do not have a common substring".format(str1,str2)

print(findstr(s1,s2))
print(findstr(s3,s2))
print(findstr(s4,s2))
print(findstr(s5,s2))
print(findstr(s6,s2))

　　输出结果：

count=2  subStringLen:6    #findstr(s1,s2)
abcdef

count=15  subStringLen:1   #findstr(s3,s2)
a

count=37  subStringLen:1    #findstr(s4,s2)
w

count=1  subStringLen:7      #findstr(s5,s2)
defabcd

'j' and 'defabcdoabcdeftw' do not have a common substring  #findstr(s6,s2)

　　

永远不要相信用户的输入，一个健壮的代码往往业务功能块代码很少，而对用户输入的验证代码部分越要更严谨，将用户所有的输入都考虑在内。