244. 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 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.

一刷
题解：与243不同的是，这个时候我们可以设计一个类，构造函数传入了array。
第二个需要注意的是，对于两个递增的数列，怎样找到i在数列a, j在数列b中的元素的最小差值。time complexity: O(m+n)

for(int i=0, j=0; i

public class WordDistance {

    //the element in the value is ascending
    private Map> map;
    
    public WordDistance(String[] words) {
        map = new HashMap<>();
        for(int i=0; i list = new ArrayList<>();
                list.add(i);
                map.put(w, list);
            }
        }
    }
    
    public int shortest(String word1, String word2) {
        List list1 = map.get(word1);
        List list2 = map.get(word2);
        int res = Integer.MAX_VALUE;
        for(int i=0, j=0; i