Hihocode Trie树代码

来源：互联网发布：淘宝super制造编辑：程序博客网时间：2024/05/17 00:04

今天AC了HiHocode 的第二道题，因为hihocode并不给测试提示，所以难度感觉leetcode高一些，比较恶心。

但是当出现AC的时候真是太高兴了，废话不多说，进入正题

我是华丽丽的分割线！

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hihocode很善意的提供了提示，虽然分了三个，但是关键点还是两个

一是如何构建树

二是如何查询前置单词

Trie树是以边来保存字符的，博主一开始误以为是节点保存了，耗费了一点脑力。

另外，hihocode也提示了需要边插入边统计，于是我们的构造体就这样出现了。

需要注意的是hihocode 定义重复的单词算两个，博主一开始去掉了重复单词，于是WA了很久。

struct Node{bool isWord;Node *child[Max];//save edgeint L[Max];//save countNode(){isWord = false;for (int i = 0; i < Max; i++){child[i] = NULL;}for (int i = 0; i < Max; i++){L[i] = 0;}}~Node(){for (int i = 0; i < Max; i++){delete child[i];}}};

插入单词：

void Insert(string word){Node *tempRoot = root;for (int i = 0; i < word.length(); i++){tempRoot->L[word[i]-'a']++;if (tempRoot->child[word[i]-'a'] != NULL){tempRoot = tempRoot->child[word[i]-'a'];}else{Node *newNode = new Node();tempRoot->child[word[i]-'a'] = newNode;tempRoot = tempRoot->child[word[i]-'a'];}if (i == word.length() - 1){if (tempRoot->isWord == false){tempRoot->isWord = true;}}}}

因为是一边查找一边统计，所以其实可以不完全符合trie树也可以，是单词的那个节点不做标记也无法影响结果

查找单词/前缀数量：

int find(string word){Node *tempRoot = root;for (int i = 0; i < word.length(); i++){if (tempRoot->child[word[i]-'a'] == NULL)return 0;else{if (i == word.length() - 1) return tempRoot->L[word[i]-'a'];tempRoot = tempRoot->child[word[i]-'a'];}}}

接下来就贴上所有的代码

所有的代码中包含减去重复单词的函数，但是并没有用到：

#include<iostream>#include<string>using namespace std;const int Max = 26;struct Node{bool isWord;Node *child[Max];//save edgeint L[Max];//save countNode(){isWord = false;for (int i = 0; i < Max; i++){child[i] = NULL;}for (int i = 0; i < Max; i++){L[i] = 0;}}~Node(){for (int i = 0; i < Max; i++){delete child[i];}}};Node * root = new Node();void minusDuplicate(string word){Node *tempRoot = root;for (int i = 0; i < word.length(); i++){tempRoot->L[word[i]-'a']--;tempRoot = tempRoot->child[word[i]-'a'];}}void Insert(string word){Node *tempRoot = root;for (int i = 0; i < word.length(); i++){tempRoot->L[word[i]-'a']++;if (tempRoot->child[word[i]-'a'] != NULL){tempRoot = tempRoot->child[word[i]-'a'];}else{Node *newNode = new Node();tempRoot->child[word[i]-'a'] = newNode;tempRoot = tempRoot->child[word[i]-'a'];}if (i == word.length() - 1){if (tempRoot->isWord == false){tempRoot->isWord = true;}}}}int find(string word){Node *tempRoot = root;for (int i = 0; i < word.length(); i++){if (tempRoot->child[word[i]-'a'] == NULL)return 0;else{if (i == word.length() - 1) return tempRoot->L[word[i]-'a'];tempRoot = tempRoot->child[word[i]-'a'];}}}int main(){int count = 0;cin >> count;string insertStr;for (int i = count; i > 0; i--){cin >> insertStr;Insert(insertStr);}cin >> count;for (int i = count; i > 0; i--){cin >> insertStr;cout<<find(insertStr)<<endl;}return 0;}

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