leetcode244- Shortest Word Distance II- medium

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.

 

1. 建立和遍历时间复杂度都是O(n)的解法。比之前的方法的优化点是取两个下标list来获得两者pair的差的最小值的求法。现在改为用两个指针各指着某一个下标，之后哪个指着的下标是小一点的，就挪哪一个指针。这样总能保证每次尽量让两个下标靠的最近，最后打擂台出来的肯定能打到最小值。

2.初始化建完全映射的一个解法，建立的时候时间复杂度是O(n4)，但是之后取的是O(1)，问要不要这个权衡吧。

用的数据结构是map<String, List<Integer>>(映射字符串和它的下标)，以及int[i][j](记录words[i]和words[j]之间的距离)。

这样之后调用的时候只要读一次map以及读一次array就好。

1.O(n), O(n)

class WordDistance {

    private Map<String, List<Integer>> map;
    
    public WordDistance(String[] words) {
        map = new HashMap<String, List<Integer>>();
        for (int i = 0; i < words.length; i++) {
            if (!map.containsKey(words[i])) {
                map.put(words[i], new ArrayList<Integer>());
            }
            map.get(words[i]).add(i);
        }
    }
    
    public int shortest(String word1, String word2) {
        List<Integer> l1 = map.get(word1);
        List<Integer> l2 = map.get(word2);
        
        int p1 = 0, p2 = 0;
        int min = Integer.MAX_VALUE;
        while (p1 < l1.size() && p2 < l2.size()) {
            min = Math.min(min, Math.abs(l1.get(p1) - l2.get(p2)));
            if (l1.get(p1) <= l2.get(p2)) {
                p1++;
            } else {
                p2++;
            }
        }
        return min;
    }
}

/**
 * Your WordDistance object will be instantiated and called as such:
 * WordDistance obj = new WordDistance(words);
 * int param_1 = obj.shortest(word1,word2);
 */

 

 

2. O(n4), O(1)

class WordDistance {

    private int[][] dist;
    private Map<String, List<Integer>> map;
    
    public WordDistance(String[] words) {
        
        dist = new int[words.length][words.length];
        map = new HashMap<String, List<Integer>>();
        
        for (int i = 0; i < words.length; i++) {
            if (!map.containsKey(words[i])) {
                map.put(words[i], new ArrayList<Integer>());
            } 
            map.get(words[i]).add(i);
        }
        
        for (int i = 0; i < words.length; i++) {
            for (int j = 0; j < words.length; j++) {
                dist[i][j] = -1;
            }
        }
        for (int i = 0; i < words.length; i++) {
            for (int j = 0; j < words.length; j++) {
                if (dist[i][j] != -1) {
                    continue;
                }
                List<Integer> idxs1 = map.get(words[i]);
                List<Integer> idxs2 = map.get(words[j]);
                int min = Integer.MAX_VALUE;
                for (int m = 0; m < idxs1.size(); m++) {
                    for (int n = 0; n < idxs2.size(); n++) {
                        min = Math.min(min, Math.abs(idxs1.get(m) - idxs2.get(n)));
                    }
                }
                for (int m = 0; m < idxs1.size(); m++) {
                    for (int n = 0; n < idxs2.size(); n++) {
                        dist[idxs1.get(m)][idxs2.get(n)] = min;
                        dist[idxs2.get(n)][idxs1.get(m)] = min;
                    }
                }
                
            }
        } 
    }
    
    public int shortest(String word1, String word2) {
        return dist[map.get(word1).get(0)][map.get(word2).get(0)];
    }
}

/**
 * Your WordDistance object will be instantiated and called as such:
 * WordDistance obj = new WordDistance(words);
 * int param_1 = obj.shortest(word1,word2);
 */