[leetcode] 244. Shortest Word Distance II 解题报告

题目链接: https://leetcode.com/problems/shortest-word-distance-ii/

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 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: 每一个单词可能出现多次, 然后将其位置存在以其字符串为key的hash表中, 然后遍历两个字符串的各个位置, 找出最短距离.

代码如下:

class WordDistance {
public:
    WordDistance(vector<string>& words) {
        for(int i =0; i< words.size(); i++)
            mp[words[i]].push_back(i);
    }
    int shortest(string word1, string word2) {
        int ans = INT_MAX;
        for(auto val1: mp[word1])
            for(auto val2: mp[word2])
                ans = min(ans, abs(val1-val2));
        return ans;
    }
private:
    unordered_map<string, vector<int>> mp;
};


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