Leetcode: 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.

Copy

Hash table method

Complexity

Time O(N) Space O(N)

Ideas

Because... Will be called many times , We can't find the subscripts of these two words every time we call . We can use a hash table , When passing in a string array , Use HashMap To store word as well as word stay Array What's in it index. So when the shortest distance method is called , We just need to traverse the subscript list of two words . Specific comparison methods , It's like merge two list, Compare two at a time list The two smallest values , You get a difference . Then remove the smaller one . Because we traverse the input array from front to back , So the subscript list is also orderly .

 1 public class WordDistance {
 2     
 3     HashMap<String, ArrayList<Integer>> map;
 4 
 5     public WordDistance(String[] words) {
 6         this.map = new HashMap<String, ArrayList<Integer>>();
 7         for (int i=0; i<words.length; i++) {
 8             String item = words[i];
 9             if (map.containsKey(item)) {
10                 map.get(item).add(i);
11             }
12             else {
13                 ArrayList<Integer> list = new ArrayList<Integer>();
14                 list.add(i);
15                 map.put(item, list);
16             }
17         }
18     }
19 
20     public int shortest(String word1, String word2) {
21         ArrayList<Integer> l1 = map.get(word1);
22         ArrayList<Integer> l2 = map.get(word2);
23         int minDis = Integer.MAX_VALUE;
24         int i=0, j=0;
25         while (i<l1.size() && j<l2.size()) {
26             int p1 = l1.get(i);
27             int p2 = l2.get(j);
28             minDis = Math.min(minDis, Math.abs(p1-p2));
29             if (p1 < p2) i++;
30             else j++;
31         }
32         return minDis;
33     }
34 }
35 
36 // Your WordDistance object will be instantiated and called as such:
37 // WordDistance wordDistance = new WordDistance(words);
38 // wordDistance.shortest("word1", "word2");
39 // wordDistance.shortest("anotherWord1", "anotherWord2");

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