[Swift]LeetCode244.最短单词距离 II $ Shortest Word Distance II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-pgqpxffu-mb.html 
➤若是连接不是山青咏芝的博客园地址，则多是爬取做者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持做者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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.

---

这是最短单词距离的后续行动。惟一的区别是，如今您获得了单词列表，而且您的方法将使用不一样的参数重复调用屡次。您将如何优化它？

设计一个类，该类接收构造函数中的单词列表，并实现一个方法，该方法接受单词1和单词2，并返回列表中这两个单词之间的最短距离。

例如，

假设words=[“practice”、“makes”、“perfect”、“coding”、“makes”]。

给定word1=“coding”，word2=“practice”，返回3。

给定word1=“makes”，word2=“coding”，返回1。

注：

您能够假定word1不等于word2，word1和word2都在列表中。

---

 1 class WordDistance {
 2     var m:[String:[Int]] = [String:[Int]]()
 3     init(_ words: [String]) {
 4         // perform some initialization here
 5         for i in 0..<words.count
 6         {
 7             //判断是否为空
 8             if m[words[i]] == nil
 9             {
10                 m[words[i]] = [Int]()
11             }
12             m[words[i]]!.append(i)
13         }
14     }
15     
16     func shortest(_ word1:String,_ word2:String) -> Int {
17         var i:Int = 0
18         var j:Int = 0
19         var res:Int = Int.max
20         while(i < m[word1]!.count && j < m[word2]!.count)
21         {
22             res = min(res, abs(m[word1]![i] - m[word2]![j]))
23             if m[word1]![i] < m[word2]![j]
24             {
25                 i += 1
26             }
27             else
28             {
29                 j += 1
30             }
31         }
32         return res
33     }
34 }