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.

Solution：Map>

Time Complexity: 初始化:O(N) shortest(): O(重复个数)
Space Complexity: O(N)

Solution Code：

public class WordDistance {

    private Map> map;

    public WordDistance(String[] words) {
        map = new HashMap>();
        for(int i = 0; i < words.length; i++) {
            String w = words[i];
            if(map.containsKey(w)) {
                map.get(w).add(i);
            } else {
                List 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 ret = Integer.MAX_VALUE;
        for(int i = 0, j = 0; i < list1.size() && j < list2.size(); ) {
            int index1 = list1.get(i), index2 = list2.get(j);
            if(index1 < index2) {
                ret = Math.min(ret, index2 - index1);
                i++;
            } else {
                ret = Math.min(ret, index1 - index2);
                j++;
            }
        }
        return ret;
    }
}