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.
一刷
題解：與243不同的是，這個(gè)時(shí)候我們可以設(shè)計(jì)一個(gè)類，構(gòu)造函數(shù)傳入了array。
第二個(gè)需要注意的是褂策，對于兩個(gè)遞增的數(shù)列，怎樣找到i在數(shù)列a, j在數(shù)列b中的元素的最小差值旺聚。time complexity: O(m+n)
for(int i=0, j=0; i<list1.size() && j<list2.size();){
int index1 = list1.get(i), index2 = list2.get(j);
if(index1<index2){
res = Math.min(res, index2-index1);
i++;
}else{
res = Math.min(res, index1-index2);
j++;
}
}
public class WordDistance {
//the element in the value is ascending
private Map<String, List<Integer>> map;
public WordDistance(String[] words) {
map = new HashMap<>();
for(int i=0; i<words.length; i++){
String w = words[i];
if(map.containsKey(w)) map.get(w).add(i);
else{
List<Integer> list = new ArrayList<>();
list.add(i);
map.put(w, list);
}
}
}
public int shortest(String word1, String word2) {
List<Integer> list1 = map.get(word1);
List<Integer> list2 = map.get(word2);
int res = Integer.MAX_VALUE;
for(int i=0, j=0; i<list1.size() && j<list2.size();){
int index1 = list1.get(i), index2 = list2.get(j);
if(index1<index2){
res = Math.min(res, index2-index1);
i++;
}else{
res = Math.min(res, index1-index2);
j++;
}
}
return res;
}
}
/**
* Your WordDistance object will be instantiated and called as such:
* WordDistance obj = new WordDistance(words);
* int param_1 = obj.shortest(word1,word2);
*/
