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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。
复制代码
- 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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