力扣208. 实现 Trie (前缀树)

原题

以下是我的代码，就是简单的字符串操作，可以ac但背离了题意，我之前没接触过Trie 

class Trie:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.trie = set()

    def insert(self, word: str) -> None:
        """
        Inserts a word into the trie.
        """
        self.trie.add(word)

    def search(self, word: str) -> bool:
        """
        Returns if the word is in the trie.
        """
        return word in self.trie

    def startsWith(self, prefix: str) -> bool:
        """
        Returns if there is any word in the trie that starts with the given prefix.
        """
        lens = len(prefix)
        for s in self.trie:
            if s[:lens] == prefix:
                return True
        return False


# Your Trie object will be instantiated and called as such:
# obj = Trie()
# obj.insert(word)
# param_2 = obj.search(word)
# param_3 = obj.startsWith(prefix)

以下是大佬写的，答案链接

class Trie {
private:
    bool isEnd;
    Trie* next[26];
public:
    Trie() {
        isEnd = false;
        memset(next, 0, sizeof(next));
    }
    
    void insert(string word) {
        Trie* node = this;
        for (char c : word) {
            if (node->next[c-'a'] == NULL) {
                node->next[c-'a'] = new Trie();
            }
            node = node->next[c-'a'];
        }
        node->isEnd = true;
    }
    
    bool search(string word) {
        Trie* node = this;
        for (char c : word) {
            node = node->next[c - 'a'];
            if (node == NULL) {
                return false;
            }
        }
        return node->isEnd;
    }
    
    bool startsWith(string prefix) {
        Trie* node = this;
        for (char c : prefix) {
            node = node->next[c-'a'];
            if (node == NULL) {
                return false;
            }
        }
        return true;
    }
};