LeetCode 208 Implement Trie

LeetCode 208 Implement Trie

思路

这一题的关键在于树的节点如何实现。

我这里采用isWord表示到达该节点的路径是否组成单词。例如，插入单词appple后，节点e的isWord变量为true，第二个p的isWord为false。接着再插入单词app，则第二个p的isWord改为true即可。

通过一个拥有26个元素的vector保存子节点。在构造函数中，全部初始化为nullptr。在析构函数中，将非nullptr节点删除。

代码

struct TrieNode {
    bool isWord;
    vector<TrieNode*> next;
    TrieNode(): isWord(false), next(26, nullptr) { }
    ~TrieNode() {
        for (auto &node : next)
            if (node) {
                delete node;
                node = nullptr;
            }
    }
};

class Trie {
public:
    /** Initialize your data structure here. */
    Trie() {
        root = new TrieNode();
    }
    
    /** Inserts a word into the trie. */
    void insert(string word) {
        TrieNode *p = root;
        for (int i = 0; i < word.size(); ++i) {
            int oft = word[i] - 'a';
            if (!p->next[oft])
                p->next[oft] = new TrieNode();
            p = p->next[oft];
        }
        p->isWord = true;
    }
    
    /** Returns if the word is in the trie. */
    bool search(string word) {
        TrieNode *p = root;
        for (int i = 0; i < word.size(); ++i) {
            int oft = word[i] - 'a';
            if (!p->next[oft])
                return false;
            p = p->next[oft];
        }
        return p->isWord;
    }
    
    /** Returns if there is any word in the trie that starts with the given prefix. */
    bool startsWith(string prefix) {
        TrieNode *p = root;
        for (int i = 0; i < prefix.size(); ++i) {
            int oft = prefix[i] - 'a';
            if (!p->next[oft])
                return false;
            p = p->next[oft];
        }
        return true;
    }
    
    ~Trie() { delete root; }

    private:
        TrieNode *root;
    
};
    
    /**
     * Your Trie object will be instantiated and called as such:
     * Trie* obj = new Trie();
     * obj->insert(word);
     * bool param_2 = obj->search(word);
     * bool param_3 = obj->startsWith(prefix);
     */