LeetCode Shortest Word Distance II

原题链接在这里：https://leetcode.com/problems/shortest-word-distance-ii/

类似Shortest Word Distance.

method 会被call好多次，没有必要每次都traverse 一遍words array. 简历一个HashMap, key是word, value是该word出现index的list.

比较word1 和 word2 两个index list. 跟新最小距离.

AC Java:

 1 public class WordDistance {
 2     HashMap<String,List<Integer>> hm;
 3     public WordDistance(String[] words) {
 4         hm = new HashMap<String,List<Integer>>();
 5         for(int i = 0; i<words.length; i++){
 6             if(!hm.containsKey(words[i])){
 7                 List<Integer> item = new ArrayList<Integer>();
 8                 item.add(i);
 9                 hm.put(words[i],item);
10             }else{
11                 hm.get(words[i]).add(i);
12             }
13         }
14     }
15 
16     public int shortest(String word1, String word2) {
17         List<Integer> l1 = hm.get(word1);
18         List<Integer> l2 = hm.get(word2);
19         int res = Integer.MAX_VALUE;
20         for(int i : l1){
21             for(int j : l2){
22                 res = Math.min(res, Math.abs(j-i));
23             }
24         }
25         return res;
26     }
27 }
28 
29 // Your WordDistance object will be instantiated and called as such:
30 // WordDistance wordDistance = new WordDistance(words);
31 // wordDistance.shortest("word1", "word2");
32 // wordDistance.shortest("anotherWord1", "anotherWord2");