面试题 17.11. 单词距离

题目

有个内含单词的超大文本文件，给定任意两个单词，找出在这个文件中这两个单词的最短距离(相隔单词数)。如果寻找过程在这个文件中会重复多次，而每次寻找的单词不同，你能对此优化吗?

示例：

输入：words = ["I","am","a","student","from","a","university","in","a","city"], word1 = "a", word2 = "student"
输出：1

分析： 一种解法是采用双指针，遍历数组，记录找到的单词的位置，两个指针单词不同则计算中间的距离，相同则更新前指针位置到后指针，后指针继续遍历。第二种方法是分别记录下两个单词在文件中出现的所有坐标，然后计算两个坐标组之间的最小距离。
方法1： 双指针

class Solution:
    def findClosest(self, words: List[str], word1: str, word2: str) -> int:
        word1_loc = []
        word2_loc = []
        for i in range(len(words)):
            if words[i] == word1:
                word1_loc.append(i)
            elif words[i] == word2:
                word2_loc.append(i)
        i, j = 0, 0
        min_distance = len(words)
        while i < len(word1_loc) and j < len(word2_loc):
            min_distance = min(min_distance, abs(word1_loc[i]-word2_loc[j]))
            if word1_loc[i] > word2_loc[j]:
                j += 1
            else:
                i += 1
        return min_distance

方法2：

class Solution:
    def findClosest(self, words: List[str], word1: str, word2: str) -> int:
        word1_loc = []
        word2_loc = []
        for i in range(len(words)):
            if words[i] == word1:
                word1_loc.append(i)
            elif words[i] == word2:
                word2_loc.append(i)
        i, j = 0, 0
        min_distance = len(words)
        while i < len(word1_loc) and j < len(word2_loc):
            min_distance = min(min_distance, abs(word1_loc[i]-word2_loc[j]))
            if word1_loc[i] > word2_loc[j]:
                j += 1
            else:
                i += 1
        return min_distance