[LeetCode] 243. Shortest Word Distance_Easy

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

Example:
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Input: word1 = “coding”, word2 = “practice”
Output: 3

Input: word1 = "makes", word2 = "coding"
Output: 1

Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.

基本思路为O(n^2) 就是每次扫了word1之后再扫word2.

Imporve 思路为O(n), 也就是每次找到两个相应的最近的点, 我们就跟ans比较一下, ans的初始化为len(words)

Code

class Solution:
    def shortestDistance(self, words, word1,word2):
        point1, point2, ans, words = None, None, len(words), [0] + words # 有[0]是为了方便edge case, 我们直接用point1 and point2来判断, 否则point1 or point2 == 0 的时候会影响        for i in range(1, len(words)):
            if words[i] == word1:
                point1 = i
            if words[i] == word2:
                point2 = i
            if point1 and point2:
                ans = min(ans, abs(point1 - point2))
        return ans