设计一个方法,找出任意指定单词在一本书中的出现频率。
你的实现应该支持如下操作:
WordsFrequency(book)
构造函数,参数为字符串数组构成的一本书get(word)
查询指定单词在数中出现的频率示例:
WordsFrequency wordsFrequency = new WordsFrequency(
{"i", "have", "an", "apple", "he", "have", "a", "pen"});
wordsFrequency.get("you"); //返回0,"you"没有出现过
wordsFrequency.get("have"); //返回2,"have"出现2次
wordsFrequency.get("an"); //返回1
wordsFrequency.get("apple"); //返回1
wordsFrequency.get("pen"); //返回1
提示:
book[i]中只包含小写字母
1 <= book.length <= 100000
1 <= book[i].length <= 10
get函数的调用次数不会超过100000
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/words-frequency-lcci 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
class WordsFrequency {
unordered_map<string,int> m;
public:
WordsFrequency(vector<string>& book) {
for(auto& s : book)
m[s]++;
}
int get(string word) {
return m[word];
}
};
class Trie
{
public:
unordered_map<char,Trie*> next;
bool isEnd = false;
int count = 0;
void insert(string& s)
{
Trie *root = this;
for(char ch : s)
{
if(!(root->next).count(ch))
{
Trie* node = new Trie();
root->next.insert(make_pair(ch,node));
}
root = root->next[ch];
}
root->isEnd = true;
root->count++;
}
int search(string& s)
{
Trie * root = this;
for(char ch : s)
{
if(!(root->next).count(ch))
{
return 0;
}
root = root->next[ch];
}
if(root->isEnd)
return root->count;
return 0;
}
};
class WordsFrequency {
Trie *t;
public:
WordsFrequency(vector<string>& book) {
t = new Trie();
for(string& b : book)
t->insert(b);
}
int get(string word) {
return t->search(word);
}
};