Implement Magic Dictionary

删除给定字符串的一个字符以后，可否在字典中找到删除一个字符以后相等的值

问题：

Implement a magic directory with buildDict, and search methods.

For the method buildDict, you'll be given a list of non-repetitive words to build a dictionary.

For the method search, you'll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:

Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False

Note:

1. You may assume that all the inputs are consist of lowercase letters a-z.
2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
3. Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.

解决：

①  

• 对于字典中的每一个单词，每一个字符，删除字符并将单词的其他部分做为关键字，将删除的字符和字符的下标做为值列表的一部分放入map中。
• 例如：“hello” -> {“ello”:[[0, ‘h’]], “hllo”:[[1, ‘e’]], “helo”:[[2, ‘l’],[3, ‘l’]], “hell”:[[4, ‘o’]]}
• 在查找过程当中，如步骤1所示生成key。当键值对中有一对相同的index但char不一样时，返回true。 例如“healo”当删除一个键时，键是“helo”，而且有一个索引相同但char不一样的[2，'l']，则返回true。

class MagicDictionary { //141ms
    Map<String,List<int[]>> map = new HashMap<>();
    /** Initialize your data structure here. */
    public MagicDictionary() {}
    /** Build a dictionary through a list of words */
    public void buildDict(String[] dict) {
        for (String s : dict){
            for (int i = 0;i < s.length();i ++){
                String key = s.substring(0,i) + s.substring(i + 1);
                int[] pair = new int[]{i,s.charAt(i)};
                List<int[]> val = map.getOrDefault(key,new ArrayList<>());
                val.add(pair);
                map.put(key,val);
            }
        }
    }
    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
    public boolean search(String word) {
        for (int i = 0;i < word.length();i ++){
            String key = word.substring(0,i) + word.substring(i + 1);
            if (map.containsKey(key)){
                for (int[] pair : map.get(key)){
                    if (pair[0] == i && pair[1] != word.charAt(i)){
                        return true;
                    }
                }
            }
        }
        return false;
    }
}
/**
 * Your MagicDictionary object will be instantiated and called as such:
 * MagicDictionary obj = new MagicDictionary();
 * obj.buildDict(dict);
 * boolean param_2 = obj.search(word);
 */ 

② 使用前缀树实现。

class MagicDictionary { //106ms     class TrieNode{         TrieNode[] children = new TrieNode[26];         boolean isWord = false;     }     TrieNode root;     /** Initialize your data structure here. */     public MagicDictionary() {         root = new TrieNode();     }     /** Build a dictionary through a list of words */     public void buildDict(String[] dict) {         for (String word : dict){             TrieNode cur = root;             for (char c : word.toCharArray()){                 int i = c - 'a';                 if (cur.children[i] == null) cur.children[i] = new TrieNode();                 cur = cur.children[i];             }             cur.isWord = true;         }     }     /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */     public boolean search(String word) {         return search(root,word.toCharArray(),0,0);     }     public boolean search(TrieNode root,char[] word,int start,int changed){         if (start == word.length) return root.isWord && changed == 1;         int index = word[start] - 'a';         for (int i = 0;i < 26;i ++){             if (root.children[i] == null) continue;             if (i == index){                 if (search(root.children[i],word,start + 1,changed)) return true;             }else {                 if (changed == 0 && search(root.children[i],word,start + 1,changed + 1)) return true;             }         }         return false;     } } /**  * Your MagicDictionary object will be instantiated and called as such:  * MagicDictionary obj = new MagicDictionary();  * obj.buildDict(dict);  * boolean param_2 = obj.search(word);  */