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。
参考:Trie树
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
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