LeetCode 244. Shortest Word Distance II（最短单词距离）

原题网址：https://leetcode.com/problems/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.
思路：先记录每个单词在数组中的位置，然后对两个单词的所有位置顺序扫描。

public class WordDistance {
private Map<String, List<Integer>> map = new HashMap<>();
public WordDistance(String[] words) {
for(int i=0; i<words.length; i++) {
List<Integer> positions = map.get(words[i]);
if (positions == null) {
positions = new ArrayList<>();
map.put(words[i], positions);
}
positions.add(i);
}
}

public int shortest(String word1, String word2) {
int distance = Integer.MAX_VALUE;
List<Integer> positions1 = map.get(word1);
List<Integer> positions2 = map.get(word2);
int i=0, j=0;
while (i<positions1.size() && j<positions2.size()) {
if (positions1.get(i) < positions2.get(j)) {
if (positions2.get(j)-positions1.get(i) < distance) distance = positions2.get(j)-positions1.get(i);
i ++;
} else {
if (positions1.get(i)-positions2.get(j) < distance) distance = positions1.get(i)-positions2.get(j);
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");