LeetCode 题解(268) : Shortest Word Distance II

最新推荐文章于 2019-11-07 13:39:00 发布

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最新推荐文章于 2019-11-07 13:39:00 发布

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分类专栏：算法文章标签： algorithm leetcode 面试题

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本文介绍了一种通过预处理和存储单词位置来优化查询效率的方法，实现快速查找两个指定单词在列表中的最短距离。利用Python实现了一个名为WordDistance的类，该类在初始化时接收单词列表并记录每个单词出现的所有位置，从而在多次查询中实现高效响应。

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题目：

This is a follow up of Shortest Word Distance. The only difference is now you are given the list of words and your method will be calledrepeatedly many times with different parameters. How would you optimize it?

Design a class which receives a list of words in the constructor, and implements a method that takes two wordsword1 and word2 and return the shortest distance between these two words in the list.

For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = “coding”, word2 = “practice”, return 3.
Given word1 = "makes", word2 = "coding", return 1.

Note:
You may assume that word1 does not equal to word2, andword1 and word2 are both in the list.

题解：

和上题解法一样。

Python版：

import sys

class WordDistance(object):
    def __init__(self, words):
        """
        initialize your data structure here.
        :type words: List[str]
        """
        self.d = {}
        for i in range(len(words)):
            if words[i] in self.d:
                self.d[words[i]].append(i)
            else:
                self.d[words[i]] = [i]

    def shortest(self, word1, word2):
        """
        Adds a word into the data structure.
        :type word1: str
        :type word2: str
        :rtype: int
        """
        s= sys.maxint
        for i in self.d[word1]:
            for j in self.d[word2]:
                a = abs(i - j)
                if a < s:
                    s = a
        return s


# Your WordDistance object will be instantiated and called as such:
# wordDistance = WordDistance(words)
# wordDistance.shortest("word1", "word2")
# wordDistance.shortest("anotherWord1", "anotherWord2")