剑指offer专项突击版第21天

剑指 Offer II 062. 实现前缀树

自己写的版本，内存消耗击败5% 191MB。不明白什么原因，官方的思路是差不多的。

$idx$ 表示的是某个字符串

class Trie {
public:
    int son[100010][26]{0}, idx;
    bool st[100010]{0};
    /** Initialize your data structure here. */
    Trie() {
        idx= 0;
    }
    
    /** Inserts a word into the trie. */
    void insert(string word) {
        int p = 0;
        for(int i = 0; i < word.size(); i++) {
            int u = word[i] - 'a';
            if(!son[p][u]) son[p][u] = ++idx;
            p = son[p][u];
        }
        st[p] = true;
    }
    
    /** Returns if the word is in the trie. */
    bool search(string word) {
        int p = 0;
        for(int i = 0; i < word.size(); i++) {
            int u = word[i]-'a';
            if(!son[p][u]) return false;
            p = son[p][u];
        }
        return st[p];
    }
    
    /** Returns if there is any word in the trie that starts with the given prefix. */
    bool startsWith(string prefix) {
        int p = 0;
        for(int i = 0; i < prefix.size(); i++) {
            int u = prefix[i]-'a';
            if(!son[p][u]) return false;
            p = son[p][u];
        }
        return true;
    }
};
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官解，内存消耗47.2MB
每个Trie类对象都表示一个结点。

class Trie {
private:
    vector<Trie*> children;
    bool isEnd;
public:
    /** Initialize your data structure here. */
    Trie():children(26),isEnd(false) {}
    
    /** Inserts a word into the trie. */
    void insert(string word) {
        auto node = this;
        for(int i = 0; i < word.size(); i++) {
            int u = word[i]-'a';
            if(!node->children[u]) {
                node->children[u] = new Trie();
            } 
            node = node->children[u];
        }
        node->isEnd = true;
    }
    
    /** Returns if the word is in the trie. */
    bool search(string word) {
        auto node = searchPrefix(word);
        return node && node->isEnd;
    }
    
    /** Returns if there is any word in the trie that starts with the given prefix. */
    bool startsWith(string prefix) {
        return searchPrefix(prefix);
    }

    Trie* searchPrefix(string word) {
        auto node = this;
        for(char c: word) {
            c -= 'a';
            if(!node->children[c]) return nullptr;
            node = node->children[c];
        }
        return node;
    }    
};
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---

剑指 Offer II 063. 替换单词

字典树

• 将 $dictionary$ 里的所有词根放到 $trie$ 上，为了更方便操作将 $sentence$ 切割放在 $vector$ 中，接着逐一去找 $word$ 的前缀是否在 $trie$ 上（搜索时只需要最短路径即可），如果有就将当前 $word$ 替换成 $trie$ 的前缀即可。最终再将所有 $word$ 拼接成答案。
• 注意：如果是以 $sentence$ 为字典会出现，$word$ 序列中相同的前缀互相替换，造成自己替换自己这就不正确了。

class Solution {
public:
    int son[100010][26]{0}, idx = 0;
    bool st[100010]{0};
    string replaceWords(vector<string>& dictionary, string sentence) {
        create_trie(dictionary);
        vector<string> words = split(sentence,' ');
        for(auto &word: words) {
            word = find_root(word);
        }
        string res;
        for(int i = 0; i < words.size() - 1; i++) {
            res.append(words[i]+" ");
        }
        res.append(words.back());
        return res;
    }

    void create_trie(vector<string> &dictionary) {
        for(string &str: dictionary) {
            int p = 0;
            for(int i = 0; i < str.size(); i++) {
                int u = str[i]-'a';
                if(!son[p][u]) son[p][u] = ++idx;
                p = son[p][u];
            }
            st[p] = true;
        }
    }

     //切割字符串
    vector<string> split(string &sentence,char ch) {
        int pos = 0, start;
        vector<string> res;
        while(pos < sentence.size()) {
            while(pos < sentence.size() && sentence[pos] == ch) {
                pos++;
            }
            start = pos;
            while(pos < sentence.size() && sentence[pos] != ch) pos++;
            if(start < sentence.size()){
                res.emplace_back(sentence.substr(start,pos-start));
            }
        }
        return res;
    }

    string find_root(string str) {
        int p = 0;
        string root;
        for(int i = 0; i < str.size(); i++) {
            int u = str[i]-'a';
            //没找到说明词根不在trie上，返回原来的字符串即可
            if(!son[p][u]) return str; 
            root += str[i];
            p = son[p][u];
            //找到最短路径的词根，直接返回
            if(st[p]) return root;
        }
        return str;
    }

};
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---

剑指 Offer II 064. 神奇的字典

由于 $buildDict$ 只调用一次，所以可以将他写入 $trie$ 字典中，然后暴力遍历 $searchWord$ 每个字符并修改，修改后查 $buildDict$ 的 $trie$ 字典。
这个方法时间效率低下。

class MagicDictionary {
private:    
struct Trie{
    vector<Trie*> children;
    bool isEnd;
    Trie():children(26),isEnd(false){} 
}*trie;
public:
    /** Initialize your data structure here. */
    MagicDictionary() {
        trie = new Trie();      
    }
    
    void buildDict(vector<string> dictionary) {
        Trie *node;
        for(auto &str: dictionary) {
            node = trie;
            for(auto ch: str) {
                ch -= 'a';
                if(!node->children[ch]) {
                    node->children[ch] = new Trie();
                }
                node = node->children[ch];
            }
            node->isEnd = true;
        }
    }
    
    bool search_word(string &word) {
        auto node = trie;
        for(int i = 0; i < word.size(); i++) {
            int u = word[i]-'a';
            if(!node->children[u]) return false;
            node = node->children[u];
        }
        return node->isEnd;
    }

    bool search(string searchWord) {
        for(int i = 0; i < searchWord.size(); i++) {
            char ch = searchWord[i];
            for(int j = 0; j < 26; j++) {
                searchWord[i] = (char)(j+'a');
                if(searchWord[i] != ch && search_word(searchWord)) return true;
            }
            searchWord[i] = ch;
        }
        return false;
    }
};

/**
 * Your MagicDictionary object will be instantiated and called as such:
 * MagicDictionary* obj = new MagicDictionary();
 * obj->buildDict(dictionary);
 * bool param_2 = obj->search(searchWord);
 */
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暴力枚举，效率还更高了，可能是数据不强。

class MagicDictionary {
private:
    vector<string> words;
    
public:
    MagicDictionary() {}
    
    void buildDict(vector<string> dictionary) {
        words = dictionary;
    }
    
    bool search(string searchWord) {
        for (auto&& word: words) {
            if (word.size() != searchWord.size()) {
                continue;
            }

            int diff = 0;
            for (int i = 0; i < word.size(); ++i) {
                if (word[i] != searchWord[i]) {
                    ++diff;
                    if (diff > 1) {
                        break;
                    }
                }
            }
            if (diff == 1) {
                return true;
            }
        }
        return false;
    }
};
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官解的字典树优化，效率比我写的高很多

struct Trie {
    bool is_finished;
    Trie* child[26];

    Trie() {
        is_finished = false;
        fill(begin(child), end(child), nullptr);
    }
};

class MagicDictionary {
public:
    MagicDictionary() {
        root = new Trie();
    }
    
    void buildDict(vector<string> dictionary) {
        for (auto&& word: dictionary) {
            Trie* cur = root;
            for (char ch: word) {
                int idx = ch - 'a';
                if (!cur->child[idx]) {
                    cur->child[idx] = new Trie();
                }
                cur = cur->child[idx];
            }
            cur->is_finished = true;
        }
    }
    
    bool search(string searchWord) {
        function<bool(Trie*, int, bool)> dfs = [&](Trie* node, int pos, bool modified) {
            if (pos == searchWord.size()) {
                return modified && node->is_finished;
            }
            int idx = searchWord[pos] - 'a';
            if (node->child[idx]) {
                if (dfs(node->child[idx], pos + 1, modified)) {
                    return true;
                }
            }
            if (!modified) {
                for (int i = 0; i < 26; ++i) {
                    if (i != idx && node->child[i]) {
                        if (dfs(node->child[i], pos + 1, true)) {
                            return true;
                        }
                    }
                }
            }
            return false;
        };

        return dfs(root, 0, false);
    }

private:
    Trie* root;
};
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