字符串递归问题

Question

I am relatively new to python and especially recursion. I have been trying to solve this problem for a while now but it's bested me so far

I have to create a function that takes in 2 strings and uses recursion to see if elements from the first string are in the second string if so it returns true if not than it returns false say "Carpool" and "prlo" or "elephant" and "xph" in the first case the recursive argument would return True and in the second case it would return False.

This is what I've been trying to do. I set a count so that if the elements match I can increase it and at the end if it's equal to len(str2) then I print 2. The major problem is iterating through the 2 strings separately, any idea on how I should approach this?

def compare(str1,str2):

count = 0
if str1 == str2:
    return "True"           
elif str1[0] == str2[0]:
    count = count + 1
    return letCheck (str1[1:],str2)
elif str1[0] != str2[0]:
    return letCheck (str1[1:],str2)

Answer 1

def compare(s1, s2):
    if not s2:
        return True
    elif s2[0] in s1:
        return compare(s1, s2[1:])
    else:
        return False

On your examples:

>>> compare( "Carpool" , "prlo")
True
>>> compare("elephant", "xph")
False

How it works

The function processes each character in string 2, s2 , one at a time. There are three cases:

1. If there are no characters left to process in s2 , that means that all the tests have passed and we return True.

2. If there are still characters left in s2 , then we test s2[0] . If it passes, we then use recursion to test the rest of the string.

3. If s2[0] is not in s1 , then we have a fail and we return False.

Answer 2

You don't need a counter. You can just check if the first letter of the first String (str1) is in the second string (str2). If so, then check if this was the last character of the first string. If it wasn't the last character, pass the rest of the string to the function again (recursion). And if one character of the first string is not in the second one, you return False.

def compare(str1, str2):
    if str1[0] in str2:
        if len(str1) == 1:
            return True;
        return compare(str[1:0], str2);
    else:
        return False;

print compare('ak', 'abcd');  # False
print compare('cd', 'abcd');  # True
print compare('efgh', 'efg'); # False   

Answer 3

This will return False if str2 is passed in as an empty string:

def letCheck(str1, str2, count = 0):
    if str1 == str2: # both strings are equal,we don't have to  go any further
        return True
    elif str2 == "" and count == 0: # if str2 is empty and count is 0 we got passed an empty str2
        return False
    elif len(str2) == count: # if count and len are equal, all str2 letters are in str1
        return True
    elif str2[0] in str1:
        count += 1 # increase count and move to the next letter in str2[1:]
        return letCheck(str1, str2[1:],count)
    return False # if we get here we have found a letter that is not in str1

When using a variable like count , you need to use it as a parameter to keep the value between each recursive call or else it will get reset to 0 each call.