[LeetCode] 676. Implement Magic Dictionary

Design a data structure that is initialized with a list of different words. Provided a string, you should determine if you can change exactly one character in this string to match any word in the data structure.

Implement the MagicDictionary class:

• MagicDictionary() Initializes the object.
• void buildDict(String[] dictionary) Sets the data structure with an array of distinct strings dictionary.
• bool search(String searchWord) Returns true if you can change exactly one character in searchWord to match any string in the data structure, otherwise returns false.

Example 1:

Input
["MagicDictionary", "buildDict", "search", "search", "search", "search"]
[[], [["hello", "leetcode"]], ["hello"], ["hhllo"], ["hell"], ["leetcoded"]]
Output
[null, null, false, true, false, false]

Explanation
MagicDictionary magicDictionary = new MagicDictionary();
magicDictionary.buildDict(["hello", "leetcode"]);
magicDictionary.search("hello"); // return False
magicDictionary.search("hhllo"); // We can change the second 'h' to 'e' to match "hello" so we return True
magicDictionary.search("hell"); // return False
magicDictionary.search("leetcoded"); // return False

Constraints:

• 1 <= dictionary.length <= 100
• 1 <= dictionary[i].length <= 100
• dictionary[i] consists of only lower-case English letters.
• All the strings in dictionary are distinct.
• 1 <= searchWord.length <= 100
• searchWord consists of only lower-case English letters.
• buildDict will be called only once before search.
• At most 100 calls will be made to search.

实现一个魔法字典。

设计一个使用单词列表进行初始化的数据结构，单词列表中的单词 互不相同 。 如果给出一个单词，请判定能否只将这个单词中一个字母换成另一个字母，使得所形成的新单词存在于你构建的字典中。

实现 MagicDictionary 类：

MagicDictionary() 初始化对象
void buildDict(String[] dictionary) 使用字符串数组 dictionary 设定该数据结构，dictionary 中的字符串互不相同
bool search(String searchWord) 给定一个字符串 searchWord ，判定能否只将字符串中 一个 字母换成另一个字母，使得所形成的新字符串能够与字典中的任一字符串匹配。如果可以，返回 true ；否则，返回 false 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/implement-magic-dictionary
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题目说了这么多，其实这就是一道字典树的题，唯一的不同是这道题允许匹配的单词有一个字母的修改。这里我们还是需要先创建字典树，然后利用DFS搜索判断。如果对字典树不熟悉，建议先做208题和211题。其他说明可以参见代码注释。

时间O(26^n) - n is word.length()

空间O(n) - 递归栈空间

Java实现

class MagicDictionary {
    class Node {
        Node[] next = new Node[26];
        boolean isEnd;
    }

    Node root = new Node();

    /** Initialize your data structure here. */
    public MagicDictionary() {

    }

    public void buildDict(String[] dictionary) {
        for (String s : dictionary) {
            Node cur = root;
            for (char c : s.toCharArray()) {
                if (cur.next[c - 'a'] == null) {
                    cur.next[c - 'a'] = new Node();
                }
                cur = cur.next[c - 'a'];
            }
            cur.isEnd = true;
        }
    }

    public boolean search(String searchWord) {
        return dfs(root, searchWord, 0, false);
    }

    private boolean dfs(Node cur, String word, int index, boolean used) {
        // 如果当前为空，则返回false
        if (cur == null) {
            return false;
        }

        // 退出条件，当搜索到单词的最后
        if (index == word.length()) {
            return cur.isEnd && used;
        }

        // normal case
        char c = word.charAt(index);
        for (int i = 0; i < 26; i++) {
            // 如果flag用过了但是当前字母不同，就continue去下一个分支找
            if (used && (c - 'a') != i) {
                continue;
            }
            // 接着DFS递归看之后的字母
            if (dfs(cur.next[i], word, index + 1, used || (c - 'a') != i)) {
                return true;
            }
        }
        return false;
    }
}

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

LeetCode 题目总结