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题目描述:给你一个产品数组 products 和一个字符串 searchWord ,products 数组中每个产品都是一个字符串。
请你设计一个推荐系统,在依次输入单词 searchWord 的每一个字母后,推荐 products 数组中前缀与 searchWord 相同的最多三个产品。如果前缀相同的可推荐产品超过三个,请按字典序返回最小的三个。
请你以二维列表的形式,返回在输入 searchWord 每个字母后相应的推荐产品的列表。
示例 1:
输入:products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"
输出:[
["mobile","moneypot","monitor"],
["mobile","moneypot","monitor"],
["mouse","mousepad"],
["mouse","mousepad"],
["mouse","mousepad"]
]
解释:按字典序排序后的产品列表是 ["mobile","moneypot","monitor","mouse","mousepad"]
输入 m 和 mo,由于所有产品的前缀都相同,所以系统返回字典序最小的三个产品 ["mobile","moneypot","monitor"]
输入 mou, mous 和 mouse 后系统都返回 ["mouse","mousepad"]
示例 2:
输入:products = ["havana"], searchWord = "havana"
输出:[["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]
示例 3:
输入:products = ["bags","baggage","banner","box","cloths"], searchWord = "bags"
输出:[["baggage","bags","banner"],["baggage","bags","banner"],["baggage","bags"],["bags"]]
示例 4:
输入:products = ["havana"], searchWord = "tatiana"
输出:[[],[],[],[],[],[],[]]
提示:
1 <= products.length <= 1000
1 <= Σ products[i].length <= 2 * 10^4
products[i] 中所有的字符都是小写英文字母。
1 <= searchWord.length <= 1000
searchWord 中所有字符都是小写英文字母。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/search-suggestions-system
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 Solution {
private:
//字典树的结构
struct Trie{
vector<int> word_id;
map<char, Trie*>next;
};
public:
vector<vector<string>> suggestedProducts(vector<string>& products, string searchWord) {
//构建字典树
int len = products.size();
Trie* root = new Trie();
for(int i = 0; i < len; i++){
string word = products[i];
Trie* node = root;
for(auto cha : word){
if((node -> next).count(cha) == 0){
node -> next[cha] = new Trie();
}
node = node -> next[cha];
(node -> word_id).push_back(i);
}
}
//搜索
vector<vector<string> >res;
Trie* node = root;
int store_ = -1;
for(int i = 0; i < searchWord.size(); i++){
vector<string>store;
if((node -> next).count(searchWord[i]) != 0){
node = node -> next[searchWord[i]];
for(auto id : (node -> word_id)){
store.push_back(products[id]);
}
sort(store.begin(), store.end());
if(store.size() > 3){
res.push_back(vector<string>(store.begin(), store.begin() +3));
}else{
res.push_back(store);
}
}else{
store_ = i;
break;
}
}
for(int i = store_; i < searchWord.size(); i++){
res.push_back(vector<string>());
}
return res;
}
};
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