两个字符串列表python之间的交集

Question

This is in reference to problem that is done in Java:
Finding the intersection between two list of string candidates .

I tried to find an equivalent solution in python:

def solution(S):
    length = len(S)
    if length == 0:
        return 1
    count = 0
    i = 0
    while i <= length:
        prefix = S[0:i:]
        suffix = S[length-i:length]
        print suffix
        if prefix == suffix:
            count += 1
        i += 1
    return count

print solution("")
print solution("abbabba")    
print solution("codility")

I know I am not doing the step in terms of suffix.

I am getting the values as 1,4,2 instead of 1,4,0

Answer 1

Your current code runs through giving the following prefix, suffix pairs for the second two examples:

a a
ab ba
abb bba
abba abba
abbab babba
abbabb bbabba
abbabba abbabba

c y
co ty
cod ity
codi lity
codil ility
codili dility
codilit odility
codility codility

I presume you are trying to return the number of times the first n characters of a string are the same as the last n . The reason the word codility is returning 2 is because you are starting the index from zero, so it is matching the empty string and the full string. Try instead:

def solution(S):
    length = len(S)
    if length == 0:
        return 1
    count = 0
    i = 1 # Don't test the empty string!
    while i < length: # Don't test the whole string! Use '<'
        prefix = S[:i] # Up to i
        suffix = S[length-i:] # length - i to the end
        print prefix, suffix
        if prefix == suffix:
            count += 1
        i += 1
    return count

print solution("")
print solution("abbabba")    
print solution("codility")

This returns 1, 2, 0.

Perhaps you were looking to test if the prefix is the same as the suffix backwards , and only test half the length of the string? In this case you should try the following:

def solution(S):
    length = len(S)
    if length == 0:
        return 1
    count = 0
    i = 1 # Don't test the empty string!
    while i <= (length + 1)/2: # Test up to halfway
        prefix = S[:i] # Up to i
        suffix = S[:length-i-1:-1] # Reverse string, up to length - i - 1
        print prefix, suffix
        if prefix == suffix:
            count += 1
        i += 1
    return count

print solution("")
print solution("abbabba")    
print solution("codility")

This returns 1, 4, 0.