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。
解法
方法一:排序 + 前缀树
题目要求在输入 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,按字典序返回最小的三个。
找前缀相同的产品,可以使用前缀树;而要返回字典序最小的三个产品,我们可以先对 products
数组进行排序,然后将排序后的数组下标存入前缀树中。
前缀树的每个节点维护以下信息:
children
:这是一个长度为 $26$ 的数组,用于存储当前节点的子节点,children[i]
表示当前节点的子节点中字符为 i + 'a'
的节点。
v
:这是一个数组,用于存储当前节点的子节点中的字符在 products
数组中的下标,最多存储三个下标。
搜索时,我们从前缀树的根节点开始,找到每一个前缀对应的下标数组,将其存入结果数组中。最后只需要将每个下标数组中的下标对应到 products
数组中即可。
时间复杂度 $O(L \times \log n + m)$,空间复杂度 $O(L)$。其中 $L$ 是 products
数组所有字符串的长度之和,而 $n$ 和 $m$ 分别是 products
数组的长度和 searchWord
的长度。
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36 | class Trie:
def __init__(self):
self.children: List[Union[Trie, None]] = [None] * 26
self.v: List[int] = []
def insert(self, w, i):
node = self
for c in w:
idx = ord(c) - ord('a')
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
if len(node.v) < 3:
node.v.append(i)
def search(self, w):
node = self
ans = [[] for _ in range(len(w))]
for i, c in enumerate(w):
idx = ord(c) - ord('a')
if node.children[idx] is None:
break
node = node.children[idx]
ans[i] = node.v
return ans
class Solution:
def suggestedProducts(
self, products: List[str], searchWord: str
) -> List[List[str]]:
products.sort()
trie = Trie()
for i, w in enumerate(products):
trie.insert(w, i)
return [[products[i] for i in v] for v in trie.search(searchWord)]
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53 | class Trie {
Trie[] children = new Trie[26];
List<Integer> v = new ArrayList<>();
public void insert(String w, int i) {
Trie node = this;
for (int j = 0; j < w.length(); ++j) {
int idx = w.charAt(j) - 'a';
if (node.children[idx] == null) {
node.children[idx] = new Trie();
}
node = node.children[idx];
if (node.v.size() < 3) {
node.v.add(i);
}
}
}
public List<Integer>[] search(String w) {
Trie node = this;
int n = w.length();
List<Integer>[] ans = new List[n];
Arrays.setAll(ans, k -> new ArrayList<>());
for (int i = 0; i < n; ++i) {
int idx = w.charAt(i) - 'a';
if (node.children[idx] == null) {
break;
}
node = node.children[idx];
ans[i] = node.v;
}
return ans;
}
}
class Solution {
public List<List<String>> suggestedProducts(String[] products, String searchWord) {
Arrays.sort(products);
Trie trie = new Trie();
for (int i = 0; i < products.length; ++i) {
trie.insert(products[i], i);
}
List<List<String>> ans = new ArrayList<>();
for (var v : trie.search(searchWord)) {
List<String> t = new ArrayList<>();
for (int i : v) {
t.add(products[i]);
}
ans.add(t);
}
return ans;
}
}
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55 | class Trie {
public:
void insert(string& w, int i) {
Trie* node = this;
for (int j = 0; j < w.size(); ++j) {
int idx = w[j] - 'a';
if (!node->children[idx]) {
node->children[idx] = new Trie();
}
node = node->children[idx];
if (node->v.size() < 3) {
node->v.push_back(i);
}
}
}
vector<vector<int>> search(string& w) {
Trie* node = this;
int n = w.size();
vector<vector<int>> ans(n);
for (int i = 0; i < w.size(); ++i) {
int idx = w[i] - 'a';
if (!node->children[idx]) {
break;
}
node = node->children[idx];
ans[i] = move(node->v);
}
return ans;
}
private:
vector<Trie*> children = vector<Trie*>(26);
vector<int> v;
};
class Solution {
public:
vector<vector<string>> suggestedProducts(vector<string>& products, string searchWord) {
sort(products.begin(), products.end());
Trie* trie = new Trie();
for (int i = 0; i < products.size(); ++i) {
trie->insert(products[i], i);
}
vector<vector<string>> ans;
for (auto& v : trie->search(searchWord)) {
vector<string> t;
for (int i : v) {
t.push_back(products[i]);
}
ans.push_back(move(t));
}
return ans;
}
};
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52 | type Trie struct {
children [26]*Trie
v []int
}
func newTrie() *Trie {
return &Trie{}
}
func (this *Trie) insert(w string, i int) {
node := this
for _, c := range w {
c -= 'a'
if node.children[c] == nil {
node.children[c] = newTrie()
}
node = node.children[c]
if len(node.v) < 3 {
node.v = append(node.v, i)
}
}
}
func (this *Trie) search(w string) [][]int {
node := this
n := len(w)
ans := make([][]int, n)
for i, c := range w {
c -= 'a'
if node.children[c] == nil {
break
}
node = node.children[c]
ans[i] = node.v
}
return ans
}
func suggestedProducts(products []string, searchWord string) (ans [][]string) {
sort.Strings(products)
trie := newTrie()
for i, w := range products {
trie.insert(w, i)
}
for _, v := range trie.search(searchWord) {
t := []string{}
for _, i := range v {
t = append(t, products[i])
}
ans = append(ans, t)
}
return
}
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