[Leetcode] 676. Implement Magic Dictionary 解题报告

魔法字典实现

最新推荐文章于 2019-05-05 23:37:45 发布

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#Leetcode #解题报告 #Trie #Hash table

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本文介绍了一个魔法字典的实现方式，包括构建字典和搜索功能。通过使用哈希表或字典树（Trie），实现了对输入单词进行精确一个字符修改后的匹配查找。

题目：

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.

思路：

在测试数据比较小的情况下，我们定义一个哈希表，用来存储所有字典中的单词。在search一个单词word的时候，我们搜索所有和它只有一个单词不一样的单词，如果找到了，就返回true，否则返回false。

但是当测试数据比较大的时候，哈希表就比较占空间了。所以我们可以构建一个字典树（Trie）。思路也还是一样的，就是搜索所有和它只有一个单词不一样的单词，如果找到了，就返回true，否则就返回false。

哈希表比较占用空间，但是搜索速度快；字典树特别节省空间，而且搜索速度也比较快，搜索一个单词的时间复杂度就是该单词的长度，一般而言可以认为是O(1)。

代码：

class MagicDictionary {
public:
    /** Initialize your data structure here. */
    MagicDictionary() {
        hash.clear();
    }
    
    /** Build a dictionary through a list of words */
    void buildDict(vector<string> dict) {
        for (auto &s : dict) {
            hash.insert(s);
        }
    }
    
    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
    bool search(string word) {
        for (int i = 0; i < word.length(); ++i) {
            char c = word[i];
            for (int j = 0; j < 26; ++j) {
                if ('a' + j != c) {
                    word[i] = 'a' + j;
                    if (hash.count(word) > 0) {
                        return true;
                    }
                }
            }
            word[i] = c;
        }
        return false;
    }
private:
    unordered_set<string> hash;
};

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