Complexity of palindrome test

问题

def is_palindrome(s):
    if len(s) <= 1:
        return True
    return s[0] == s[-1] and is_palindrome(s[1:-1])

My first thought was that the complexity is O(n) because each recursive call removes 2 characters.

But then I thought of the slice's complexity. According to https://wiki.python.org/moin/TimeComplexity, the complexity of getting a slice is O(k), where k = number of elements in the slice. In is_palindrome, k = n - 2, then k = n - 4, then n - 6, etc. so I thought the complexity would be O(n^2) because each call has an (at worst) O(n) slice and there are n calls.

Which one is correct?

回答1:

Imagine you have the classic O(n^2) algorithm: the double nested for-loop

for i in range(n):
    for j in range(n):
        #do_something

For every iteration of the outer loop, an entire iteration of the inner loop O(n) must be executed. This results in an O(n^2) runtime.

Now let's take a look at your algorithm - for every level of recursion, another O(n) algorithm (your slice) must be called - your recursive function is analogous to the outer loop, and your slice function is analogous to the inner loop.

Your recursion function is

O(n/2) => O(n)

and your slice function is

O(t) where t < n

An alternate O(n) way to decide whether a string is palindrome is to simply iterate over the string once and in each iteration check opposite ends of the list. Remember the index accesses are O(1)

for i in xrange(len(s)/2):
  if s[i] != s[(len(s)-1)-i]:
    return False
return True

来源：https://stackoverflow.com/questions/24153433/complexity-of-palindrome-test

标签

python

time-complexity

slice