[Leetcode] 244. Shortest Word Distance II 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/73939809

本文介绍了一种针对重复查询最短单词距离问题的优化算法。通过预先存储每个单词出现的位置，实现快速查找两个单词间的最短距离。适用于单词列表固定、查询需求频繁的场景。

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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 called repeatedly 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 words word1 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, and word1 and word2 are both in the list.

思路：

发现我们在上一道题目中已经将这道题目做到了最优，所以直接参考上一道题目吧，哈哈。

代码：

class WordDistance {
public:
    WordDistance(vector<string> words) {
        for (int i = 0; i < words.size(); ++i) {
            hash[words[i]].push_back(i);
        }
    }
    
    int shortest(string word1, string word2) {
        int i = 0, j = 0, ret = INT_MAX;
        while (i < hash[word1].size() && j < hash[word2].size()) {
            ret = min(ret, abs(hash[word1][i] - hash[word2][j]));
            hash[word1][i] < hash[word2][j] ? ++i : ++j;
        }
        return ret;
    }
private:
    unordered_map<string, vector<int>> hash;
};

/**
 * Your WordDistance object will be instantiated and called as such:
 * WordDistance obj = new WordDistance(words);
 * int param_1 = obj.shortest(word1,word2);
 */