676. Implement Magic Dictionary

问题描述

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:
You may assume that all the inputs are consist of lowercase letters a-z.
For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
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.

题目链接：

思路分析

实现一个magic dictionary 类，能给一些String，然后查找这些string中是否有一个，使得查找的string变化一个字符在这一些String中存在

Approach 1：Trie
Implement a trie tree which store all the strings. Then in search, change the char of search string one by and find if it is in the trie tree.

代码

class MagicDictionary {
    class TrieNode{
        TrieNode[] children = new TrieNode[26];
        boolean isWord;
        public TrieNode() {}
    }
    private 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 s : dict){
            TrieNode node = root;
            for (char ch : s.toCharArray()){
                if (node.children[ch - 'a'] == null){
                    node.children[ch - 'a'] = new TrieNode();
                }
                node = node.children[ch - 'a'];
            }
            node.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) {
        char[] arr = word.toCharArray();
        for (int i = 0; i < word.length(); i++){
            for (char ch = 'a'; ch <= 'z'; ch++){
                if (arr[i] == ch){
                    continue;
                }
                char org = arr[i];
                arr[i] = ch;
                if (searchHelper(arr)){
                    return true;
                }
                arr[i] = org;
            }
        }
        return false;
    }
    
    private boolean searchHelper(char[] arr){
        TrieNode node = root;
        for (char ch : arr){
            if (node.children[ch - 'a'] == null){
                return false;
            }
            node = node.children[ch - 'a'];
        }
        return node.isWord;
    }
}

/**
 * Your MagicDictionary object will be instantiated and called as such:
 * MagicDictionary obj = new MagicDictionary();
 * obj.buildDict(dict);
 * boolean param_2 = obj.search(word);
 */

时间复杂度：$O()$
空间复杂度：$O()$

---

反思

Approach 2: HashSet
Using a hashset to contain all the strings, and then in search, change characters one by one, find wether it is in the set.

class MagicDictionary {
    private Set<String> set;
    /** Initialize your data structure here. */
    public MagicDictionary() {
        set = new HashSet<String>();
    }
    
    /** Build a dictionary through a list of words */
    public void buildDict(String[] dict) {
        set.clear();
        for (String str : dict){
            set.add(str);
        }
    }
    
    /** 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) {
        StringBuilder sb = new StringBuilder(word);
        for (int i = 0; i < word.length(); i++){
            char temp = word.charAt(i);
            for (int j = 0; j < 26; j++){
                if ('a' + j == temp){
                    continue;
                }
                sb.setCharAt(i, (char)('a' + j));
                if (set.contains(sb.toString())){
                    return true;
                }
                sb.setCharAt(i, temp);
            }
        }
        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);
 */