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.

解题思路：

本题是Shortest Word Distance的升级版，因为需要对不同的word进行重复比较，算法的快慢就很关键，原有的方法试过，偶尔可以，Runtime达到了1870ms, 几乎已达Time Exceed 的边缘。因此本题考的是数据结构的使用，使用HashMap来存储the list of words, Key: word, Value: 是该word 对应的index ArrayList. Runtime 大幅度减小，缩小为436ms.

Java code

public class WordDistance {
private Map<String, ArrayList<Integer>> map;

public WordDistance(String[] words) {
map = new HashMap<String, ArrayList<Integer>>();
for(int i = 0; i< words.length; i++) {
String w = words[i];
if(map.containsKey(w)) {
map.get(w).add(i);
}else{
ArrayList<Integer> list = new ArrayList<Integer>();
list.add(i);
map.put(w,list);
}
}
}

public int shortest(String word1, String word2) {
ArrayList<Integer> index1 = map.get(word1);
ArrayList<Integer> index2 = map.get(word2);
int distance = Integer.MAX_VALUE;
for(int i : index1) {
for(int j : index2){
distance = Math.min(distance, Math.abs(i-j));
}
}
return distance;
}
}

// Your WordDistance object will be instantiated and called as such:
// WordDistance wordDistance = new WordDistance(words);
// wordDistance.shortest("word1", "word2");
// wordDistance.shortest("anotherWord1", "anotherWord2");

Reference：

1. https://leetcode.com/discuss/50190/java-solution-using-hashmap

2. https://leetcode.com/discuss/58041/java-solution-using-hashmap