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. 

题意：给定数组，和两个word,word,要求构造一个类，计算两者的最短距离，因为我们要给出很多组不同的两个word，你会怎么组织这个类呢？

分类：数组

解法1：在构造函数，使用hashMap来记录每个单词的位置

在计算最短距离的函数中，两个列表是已排序的，因此可使用类似MergeSort的思路求解最短距离。 

[java]  view plain  copy

1. public class WordDistance {  
2.     private HashMap<String, List<Integer>> indexer = new HashMap<String, List<Integer>>();  
3.     public WordDistance(String[] words) {  
4.         if (words == null) return;  
5.         for (int i = 0; i < words.length; i++) {  
6.             if (indexer.containsKey(words[i])) {  
7.                 indexer.get(words[i]).add(i);  
8.             } else {  
9.                 List<Integer> positions = new ArrayList<Integer>();  
10.                 positions.add(i);  
11.                 indexer.put(words[i], positions);  
12.             }  
13.         }  
14.     }  
15.   
16.     public int shortest(String word1, String word2) {  
17.         List<Integer> posList1 = indexer.get(word1);  
18.         List<Integer> posList2 = indexer.get(word2);  
19.         int i = 0, j = 0;  
20.         int diff = Integer.MAX_VALUE;  
21.         while (i < posList1.size() && j < posList2.size()) {  
22.             int pos1 = posList1.get(i), pos2 = posList2.get(j);  
23.             if (pos1 < pos2) {  
24.                 diff = Math.min(diff, pos2 - pos1);  
25.                 i++;  
26.             } else {  
27.                 diff = Math.min(diff, pos1 - pos2);  
28.                 j++;  
29.             }  
30.         }  
31.         return diff;  
32.     }  
33. }  

原文链接http://blog.youkuaiyun.com/crazy__chen/article/details/47859865