Leetcode 245. Shortest Word Distance III

[locked]
This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.
Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.
word1 and word2 may be the same and they represent two individual words in the list.
For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].
Given word1 = “makes”, word2 = “coding”, return 1.
Given word1 = "makes", word2 = "makes", return 3.
Note:
You may assume word1 and word2 are both in the list.

---

Still easy. Just divide to two cases: word1 equals word2 or not.

# Time:  O(n)
# Space: O(1)

class Solution:
    # @param {string[]} words
    # @param {string} word1
    # @param {string} word2
    # @return {integer}
    def shortestWordDistance(self, words, word1, word2):
        dist = float("inf")
        i, index1, index2 = 0, None, None
        while i < len(words):
            if words[i] == word1:
                if index1 is not None and word1 == word2:
                    dist = min(dist, abs(index1 - i))
                index1 = i
            elif words[i] == word2:
                index2 = i

            if index1 is not None and index2 is not None:
                dist = min(dist, abs(index1 - index2))
            i += 1

        return dist