1. 题目
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。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 | 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
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2. 解题
2.1 Trie树
参考:Trie树
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 | class trie { public: trie* next[26] = {NULL}; bool isend = false; int count = 0; void insert(string& s) { trie* cur = this; for(int i = 0; i < s.size(); i++) { if(cur->next[s[i]-'a'] == NULL) cur->next[s[i]-'a'] = new trie(); cur = cur->next[s[i]-'a']; } cur->count++; cur->isend = true; } }; class Solution { public: vector<vector<string>> suggestedProducts(vector<string>& products, string searchWord) { trie* t = new trie(); for(string& p : products) t->insert(p);//单词插入trie树 vector<vector<string>> ans(searchWord.size()); int i = 0, idx = 0, j = 0, n = searchWord.size(); string prefix = ""; trie* cur = t; for(i = 0; i < n; ++i) { prefix += searchWord[i]; for( ; j < prefix.size(); ++j) { if(cur->next[prefix[j]-'a'] == NULL) return ans; cur = cur->next[prefix[j]-'a']; }//找到前缀 dfs(cur,prefix,ans[i]); } return ans; } void dfs(trie* cur, string& str, vector<string>& list) { //遍历前缀以下的节点,找到单词 if(list.size() == 3) return; if(!cur) return; if(cur->isend) { int n = cur->count; while(list.size() < 3 && n--) list.push_back(str); } for(int i = 0; i < 26; ++i) { str += i+'a'; dfs(cur->next[i],str,list); str.pop_back(); } } }; |
1140 ms 68.4 MB
2.2 multiset
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | class Solution { public: vector<vector<string>> suggestedProducts(vector<string>& products, string searchWord) { multiset<string> s; for(string& p : products) s.insert(p);//插入multiset vector<vector<string>> ans(searchWord.size()); string prefix = ""; for(int i = 0; i < searchWord.size(); ++i) { prefix += searchWord[i];//输入单词前缀 auto start = s.lower_bound(prefix);//二分查找下界 for(auto it = start; it != s.end() && ans[i].size() < 3; ++it) { //把后序3个包含前缀的单词加入答案 if((*it).find(prefix) == 0) ans[i].push_back(*it); else break; } } return ans; } }; |
80 ms 24.4 MB