472. Concatenated Words 查找自己拼出来的单词 反向word break

Given an array of strings words (without duplicates), return all the concatenated words in the given list of words.

A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.

 

Example 1:

Input: words = ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats"; 
"dogcatsdog" can be concatenated by "dog", "cats" and "dog"; 
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat".

Example 2:

Input: words = ["cat","dog","catdog"]
Output: ["catdog"]

有啥是双重for循环解决不了的问题吗?没有。查找剩下的一截,算是常用模板了吧

参考:https://leetcode.com/problems/concatenated-words/discuss/541520/Java-DFS-%2B-Memoization-Clean-code

class Solution {
    public List<String> findAllConcatenatedWordsInADict(String[] words) {
        List<String> ans = new ArrayList<>();
        HashSet<String> wordSet = new HashSet<>(Arrays.asList(words));
        HashMap<String, Boolean> cache = new HashMap<>();
        for (String word : words) {
            System.out.println("整个的大word = " + word);
            if (dfs(word, wordSet, cache)) 
                ans.add(word);
        }
        return ans;
    }
    
    boolean dfs(String word, HashSet<String> wordSet, HashMap<String, Boolean> cache) {
        //已经判断过了的话,就直接返回
        if (cache.containsKey(word)) return cache.get(word);
        
        
        for (int i = 1; i < word.length(); i++) {
            //如果包括了前半截
            if (wordSet.contains(word.substring(0, i))) {
                //就计算后半截                
                String suffix = word.substring(i);
                //如果后半截在set中 or dfs能找到后半截,这个word就可以拼,返回true
                if (wordSet.contains(suffix) || dfs(suffix, wordSet, cache)) {
                    System.out.println("前半截word.substring(0, i) = " + word.substring(0, i));
                    System.out.println("后半截suffix = " + suffix);
                    System.out.println("此时这个word为true = " + word);
                    System.out.println("   ");
                    cache.put(word, true);                    
                    return true;
                }
            }
        }
        cache.put(word, false);
        return false;
    }
}

/*
一开始cat cats是缓存里的true,所以对整个的catsdogcats递归

整个的大word = cat
整个的大word = cats
整个的大word = catsdogcats
前半截word.substring(0, i) = dog
后半截suffix = cats
此时这个word为true = dogcats
   
前半截word.substring(0, i) = cats
后半截suffix = dogcats
此时这个word为true = catsdogcats
   
整个的大word = dog
整个的大word = dogcatsdog
前半截word.substring(0, i) = cats
后半截suffix = dog
此时这个word为true = catsdog
   
前半截word.substring(0, i) = dog
后半截suffix = catsdog
此时这个word为true = dogcatsdog
   
整个的大word = hippopotamuses
整个的大word = rat
整个的大word = ratcatdogcat
前半截word.substring(0, i) = dog
后半截suffix = cat
此时这个word为true = dogcat
   
前半截word.substring(0, i) = cat
后半截suffix = dogcat
此时这个word为true = catdogcat
   
前半截word.substring(0, i) = rat
后半截suffix = catdogcat
此时这个word为true = ratcatdogcat
   

*/

 

 
posted @ 2021-08-31 09:14  苗妙苗  阅读(27)  评论(0编辑  收藏  举报