题目

243. Shortest Word Distance

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解题方法

设置三个变量word1的当前索引 word1ind = -len(words)、word2的当前索引 word2ind = len(words)、最小距离distance = len(words)，遍历数组，当遇到word1或word2时更新其当前索引，并更新distance = min(distance, abs(word1ind - word2ind))，否则如果不是word1或word2，pass。最后返回distance。
时间复杂度：O(n)
空间复杂度：O(1)

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代码

class Solution:
    def shortestDistance(self, words: List[str], word1: str, word2: str) -> int:
        word1ind = -len(words)
        word2ind = len(words)
        distance = len(words)
        
        for i in range(len(words)):
            if words[i] == word1:
                word1ind = i
                distance = min(distance, abs(word1ind - word2ind))
            elif words[i] == word2:
                word2ind = i
                distance = min(distance, abs(word1ind - word2ind))
            else:
                pass
        
        return distance