Shortest Word Distance

Given a list of words and two words word1 and word2, 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.

分析：

因为word1或者word2在数组里可能有重复，所以，每个词可能会出现在不同的地方。用ArrayList记录word1(word2)出现的index, 然后找到最近的indexs。

考虑到后续问题是如果那个method多被多次call，我们可以用hashmap记录每个word出现的index.

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

    public int shortest(String word1, String word2) {
        List<Integer> pos1 = map.get(word1);
        List<Integer> pos2 = map.get(word2);

        int minDistance = Integer.MAX_VALUE;
        int i = ;
        int j = ;
        while (i < pos1.size() && j < pos2.size()) {
            int p1 = pos1.get(i);
            int p2 = pos2.get(j);
            if (p1 < p2) {
                minDistance = Math.min(minDistance, p2 - p1);
                i++;
            } else {
                minDistance = Math.min(minDistance, p1 - p2);
                j++;
            }
        }

        return minDistance;
    }
}

如果两个单词不一样，而且只被call一次，那么下面这个方法也不错。

 public class Solution {
     public int shortestDistance(String[] words, String word1, String word2) {
         int index1 = -;
         int index2 = -;
         int min = Integer.MAX_VALUE;
         for(int i = ; i < words.length; i++){
             if(words[i].equals(word1)){
                 index1 = i;
             }else if(words[i].equals(word2)){
                 index2 = i;
             }
             if(index1 != - && index2 != -){
                 min = Math.min(min, Math.abs(index1 - index2));
             }
         }
         return min;
     }
 }