三刷244. Shortest Word Distance II

L tag
这道题如果按照原题直接写，会tle, 注意到题目中的这个要求：
The only difference is now you are given the list of words and your method will be called repeatedly many times with different parameters.
所以如果我们每次有新的input parameters都得重新找index list的话就会很耗时，在words.length很大的时候就会导致超时。所以我们一开始就用map来存下单词和index list的mapping,这样每次call method都可以直接得到index list, 再按照原题的方法做就好了。

class WordDistance {
    Map> map;            
    public WordDistance(String[] words) {
        map = new HashMap<>();  
        int index = 0;
        for (String word : words){
            if (!map.containsKey(word)){
                map.put(word, new ArrayList(Arrays.asList(index++)));
            } else {
                map.get(word).add(index++);
            }
        }
    }
    
    public int shortest(String word1, String word2) {
        int shortest = Integer.MAX_VALUE;
        int i = 0;
        int j = 0;
        List indexes1 = map.get(word1);
        List indexes2 = map.get(word2);
        while (i < indexes1.size() && j < indexes2.size()){
            int dis = 0;
            if (indexes1.get(i) > indexes2.get(j)){
                dis = indexes1.get(i) - indexes2.get(j);
                j++;
            } else {
                dis = indexes2.get(j) - indexes1.get(i);
                i++;
            }
            shortest = Math.min(dis, shortest);
        }
        return shortest;
    }
}

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