剑指 Offer II 064. 神奇的字典

题目描述

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已构建的神奇字典中。

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 。

示例：

输入
inputs = ["MagicDictionary", "buildDict", "search", "search", "search", "search"]
inputs = [[], [["hello", "leetcode"]], ["hello"], ["hhllo"], ["hell"], ["leetcoded"]]
输出
[null, null, false, true, false, false]

解释
MagicDictionary magicDictionary = new MagicDictionary();
magicDictionary.buildDict(["hello", "leetcode"]);
magicDictionary.search("hello"); // 返回 False
magicDictionary.search("hhllo"); // 将第二个 'h' 替换为 'e' 可以匹配 "hello" ，所以返回 True
magicDictionary.search("hell"); // 返回 False
magicDictionary.search("leetcoded"); // 返回 False

提示：

1 <= dictionary.length <= 100
1 <= dictionary[i].length <= 100
dictionary[i] 仅由小写英文字母组成
dictionary 中的所有字符串 互不相同
1 <= searchWord.length <= 100
searchWord 仅由小写英文字母组成
buildDict 仅在 search 之前调用一次
最多调用 100 次 search

注意：本题与主站 676 题相同： https://leetcode.cn/problems/implement-magic-dictionary/

解法

方法一

Python3JavaC++Go

class MagicDictionary:
    def __init__(self):
        """
        Initialize your data structure here.
        """

    def _patterns(self, word):
        return [word[:i] + '*' + word[i + 1 :] for i in range(len(word))]

    def buildDict(self, dictionary: List[str]) -> None:
        self.words = set(dictionary)
        self.counter = Counter(p for word in dictionary for p in self._patterns(word))

    def search(self, searchWord: str) -> bool:
        for p in self._patterns(searchWord):
            if self.counter[p] > 1 or (
                self.counter[p] == 1 and searchWord not in self.words
            ):
                return True
        return False


# Your MagicDictionary object will be instantiated and called as such:
# obj = MagicDictionary()
# obj.buildDict(dictionary)
# param_2 = obj.search(searchWord)

class MagicDictionary {
    private Set<String> words;
    private Map<String, Integer> counter;

    /** Initialize your data structure here. */
    public MagicDictionary() {
        words = new HashSet<>();
        counter = new HashMap<>();
    }

    public void buildDict(String[] dictionary) {
        for (String word : dictionary) {
            words.add(word);
            for (String p : patterns(word)) {
                counter.put(p, counter.getOrDefault(p, 0) + 1);
            }
        }
    }

    public boolean search(String searchWord) {
        for (String p : patterns(searchWord)) {
            int cnt = counter.getOrDefault(p, 0);
            if (cnt > 1 || (cnt == 1 && !words.contains(searchWord))) {
                return true;
            }
        }
        return false;
    }

    private List<String> patterns(String word) {
        List<String> res = new ArrayList<>();
        char[] chars = word.toCharArray();
        for (int i = 0; i < chars.length; ++i) {
            char c = chars[i];
            chars[i] = '*';
            res.add(new String(chars));
            chars[i] = c;
        }
        return res;
    }
}

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

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

    void buildDict(vector<string> dictionary) {
        for (string word : dictionary) {
            words.insert(word);
            for (string p : patterns(word)) ++counter[p];
        }
    }

    bool search(string searchWord) {
        for (string p : patterns(searchWord)) {
            if (counter[p] > 1 || (counter[p] == 1 && !words.count(searchWord))) return true;
        }
        return false;
    }

private:
    unordered_set<string> words;
    unordered_map<string, int> counter;

    vector<string> patterns(string word) {
        vector<string> res;
        for (int i = 0; i < word.size(); ++i) {
            char c = word[i];
            word[i] = '*';
            res.push_back(word);
            word[i] = c;
        }
        return res;
    }
};

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

type MagicDictionary struct {
    words   map[string]bool
    counter map[string]int
}

/** Initialize your data structure here. */
func Constructor() MagicDictionary {
    return MagicDictionary{
        words:   make(map[string]bool),
        counter: make(map[string]int),
    }
}

func (this *MagicDictionary) BuildDict(dictionary []string) {
    for _, word := range dictionary {
        this.words[word] = true
        for _, p := range patterns(word) {
            this.counter[p]++
        }
    }
}

func (this *MagicDictionary) Search(searchWord string) bool {
    for _, p := range patterns(searchWord) {
        if this.counter[p] > 1 || (this.counter[p] == 1 && !this.words[searchWord]) {
            return true
        }
    }
    return false
}

func patterns(word string) []string {
    var res []string
    for i := 0; i < len(word); i++ {
        res = append(res, word[:i]+"."+word[i+1:])
    }
    return res
}

/**
 * Your MagicDictionary object will be instantiated and called as such:
 * obj := Constructor();
 * obj.BuildDict(dictionary);
 * param_2 := obj.Search(searchWord);
 */

剑指 Offer II 064. 神奇的字典

题目描述

解法

方法一

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