leetcode Implement Trie (Prefix Tree)

题目

Implement a trie with insert, search, and startsWith methods.

Note:
You may assume that all inputs are consist of lowercase letters a-z.
题目来自于:https://leetcode.com/problems/implement-trie-prefix-tree/

分析

实现一个字典树,字典树的相关概念见http://blog.youkuaiyun.com/u010902721/article/details/45749447

代码

class TrieNode {
public:
    // Initialize your data structure here.
    int is_last_char;
    TrieNode *next[26];
    TrieNode() {
        is_last_char = 0;
        for(int i = 0; i < 26; i++)
            next[i] = NULL;
    }
};

class Trie {
public:
    Trie() {
        root = new TrieNode();
    }

    // Inserts a word into the trie.
    void insert(string s) {
        int len = s.length();
        TrieNode* p = root;
        for(int i = 0; i < len; i++){
            int d = s.at(i) - 'a';
            if(p->next[d] == NULL){
                p->next[d] = new TrieNode();
                p->next[d]->is_last_char = 0;
            }
            p = p->next[d];
        }
        p->is_last_char = 1;
    }

    // Returns if the word is in the trie.
    bool search(string key) {
        TrieNode *p = root;
        int len = key.length();
        for(int i = 0; i < len; i++){
            int d = key.at(i) - 'a';
            if(p->next[d] == NULL)
                return false;
            else
                p = p->next[d];
        }
        if(p->is_last_char == 1)
            return true;
        else
            return false;

    }

    // Returns if there is any word in the trie
    // that starts with the given prefix.
    bool startsWith(string prefix) {
        TrieNode *p = root;
        int len = prefix.length();
        for(int i = 0; i < len; i++){
            int d = prefix.at(i) - 'a';
            if(p->next[d] == NULL)
                return false;
            else
                p = p->next[d];
        }
        return true;
    }

private:
    TrieNode* root;
};

// Your Trie object will be instantiated and called as such:
// Trie trie;
// trie.insert("somestring");
// trie.search("key");
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值