Leetcode Shortest Word Distance III

This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

word1 and word2 may be the same and they represent two individual words in the list.

For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = “makes”, word2 = “coding”, return 1.
Given word1 = "makes", word2 = "makes", return 3.

Note:
You may assume word1 and word2 are both in the list.

---

解题思路：

两种解法：

1. 参考Shortest Word Distance , 优先选择这种，简单明了。

2. 参考Shortest Word Distance II ，用hashmap， 结果run time特别慢

---

Java code

方法一

 public int shortestWordDistance(String[] words, String word1, String word2) {
        int p1 = -1, p2 = -1, distance = words.length;
        
        for(int i = 0; i<words.length; i++){
            if(words[i].equals(word1)){
                p1 = i;
                if(p1 != -1 && p2 != -1){
                    distance = (p1!=p2) ? Math.min(distance, Math.abs(p1-p2)): distance;
                }
            }
            if(words[i].equals(word2)){
                p2 = i;
                if(p1 != -1 && p2 != -1){
                    distance = (p1!=p2) ? Math.min(distance, Math.abs(p1-p2)): distance;
                }
            }
        }
        return distance;
    }

方法二

 public int shortestWordDistance(String[] words, String word1, String word2) {
        Map<String, ArrayList<Integer>> 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);
            }
        }
        int distance = Integer.MAX_VALUE;
        ArrayList<Integer> index1 = map.get(word1);
        ArrayList<Integer> index2 = map.get(word2);
          
        if(word1.equals(word2)) {
             for(int i = 0; i< index1.size()-1; i++) {
                 for(int j = i+1; j< index1.size(); j++) {
                      distance = Math.min(distance, Math.abs(index1.get(i)-index1.get(j)));
                 }
             } 
        }else {
            for(int i : index1) {
                for(int j : index2){
                    distance = Math.min(distance, Math.abs(i-j));
                }
            }
        }
        return distance;
    }

---

Reference:

1. https://leetcode.com/discuss/56772/simplest-java-solution