[LeetCode] 676. Implement Magic Dictionary 实现神奇字典

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.

实现一个神奇字典，包含buildDict和search函数。buildDict函数的功能是能把给的没有重复单词的列表建立一个字典，search函数的功能是存在和这个单词只有一个位置上的字符不同返回true，否则返回false。

Java:

class MagicDictionary {

    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<int[]>());
                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;
    }
}

Python:

class MagicDictionary(object):
    def _candidates(self, word):
        for i in xrange(len(word)):
            yield word[:i] + ‘*‘ + word[i+1:]

    def buildDict(self, words):
        self.words = set(words)
        self.near = collections.Counter(cand for word in words
                                        for cand in self._candidates(word))

    def search(self, word):
        return any(self.near[cand] > 1 or
                   self.near[cand] == 1 and word not in self.words
                   for cand in self._candidates(word))

Python:

# Time:  O(n), n is the length of the word
# Space: O(d)

import collections

class MagicDictionary(object):

    def __init__(self):
        """
        Initialize your data structure here.
        """
        _trie = lambda: collections.defaultdict(_trie)
        self.trie = _trie()

    def buildDict(self, dictionary):
        """
        Build a dictionary through a list of words
        :type dictionary: List[str]
        :rtype: void
        """
        for word in dictionary:
            reduce(dict.__getitem__, word, self.trie).setdefault("_end")

    def search(self, word):
        """
        Returns if there is any word in the trie that equals to the given word after modifying exactly one character
        :type word: str
        :rtype: bool
        """
        def find(word, curr, i, mistakeAllowed):
            if i == len(word):
                return "_end" in curr and not mistakeAllowed

            if word[i] not in curr:
                return any(find(word, curr[c], i+1, False) for c in curr if c != "_end")                            if mistakeAllowed else False

            if mistakeAllowed:
                return find(word, curr[word[i]], i+1, True) or                        any(find(word, curr[c], i+1, False)                            for c in curr if c not in ("_end", word[i]))
            return find(word, curr[word[i]], i+1, False)

        return find(word, self.trie, 0, True)　　

C++:

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

    /** Build a dictionary through a list of words */
    void buildDict(vector<string> dict) {
        for (string word : dict) {
            m[word.size()].push_back(word);
        }
    }

    /** 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 (string str : m[word.size()]) {
            int cnt = 0, i = 0;
            for (; i < word.size(); ++i) {
                if (word[i] == str[i]) continue;
                if (word[i] != str[i] && cnt == 1) break;
                ++cnt;
            }
            if (i == word.size() && cnt == 1) return true;
        }
        return false;
    }

private:
    unordered_map<int, vector<string>> m;
};　　

C++:

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

    /** Build a dictionary through a list of words */
    void buildDict(vector<string> dict) {
        for (string word : dict) s.insert(word);
    }

    /** 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.size(); ++i) {
            char t = word[i];
            for (char c = ‘a‘; c <= ‘z‘; ++c) {
                if (c == t) continue;
                word[i] = c;
                if (s.count(word)) return true;
            }
            word[i] = t;
        }
        return false;
    }

private:
    unordered_set<string> s;
};

类似题目：

[LeetCode] 208. Implement Trie (Prefix Tree) 实现字典树(前缀树)

720. Longest Word in Dictionary

原文地址：https://www.cnblogs.com/lightwindy/p/9808248.html