1268.Search-Suggestions-System

1268. Search Suggestions System

题目地址

题目描述

Given an array of strings products and a string searchWord. We want to design a system that suggests at most three product names from products after each character of searchWord is typed. Suggested products should have common prefix with the searchWord. If there are more than three products with a common prefix return the three lexicographically minimums products.
​
Return list of lists of the suggested products after each character of searchWord is typed.
​
Example 1:
​
Input: products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"
Output: [
["mobile","moneypot","monitor"],
["mobile","moneypot","monitor"],
["mouse","mousepad"],
["mouse","mousepad"],
["mouse","mousepad"]
]
​
Explanation: products sorted lexicographically = ["mobile","moneypot","monitor","mouse","mousepad"]
After typing m and mo all products match and we show user ["mobile","moneypot","monitor"]
After typing mou, mous and mouse the system suggests ["mouse","mousepad"]

代码

Approach 1:

word < word + "~"
prefix + "~" helps you find the upper bound, '~' is one option, any character after 'z' should be also working such as '{'.
class Solution {
public List<List<String>> suggestedProducts(String[] products, String searchWord) {
​
List<List<String>> res = new ArrayList<>();
if (products == null || searchWord == null) {
return res;
}
​
Arrays.sort(products);
List<String> proList = Arrays.asList(products);
TreeMap<String, Integer> proToIndex = new TreeMap<>();
for (int i = 0; i < products.length; i++) {
proToIndex.put(products[i], i);
}
​
String key = "";
for (char c : searchWord.toCharArray()) {
key = key + c;
String ceil = proToIndex.cellingKey(key);
String floor = proToIndex.floorKey(key + "~"); // can find key
if (cell == null || floor == null) {
break;
}
​
res.add(proList.subList(proToIndex.get(ceil), Math.min(proToIndex.get(ceil) + 3, proToIndex.get(floor) + 1)));
}
​
while (res.size() < searchWord.length()) {
res.add(new Array<String>());
}
​
return res;
}
}

Approach #2 Tire Tree

class Solution {
private class TrieNode {
boolean end = false;
String str = null;
int count = 0;
TrieNode[] children = new TrieNode[26];
}
​
private class Trie {
TrieNode root = new TrieNode();
public void insert(String[] products) {
for (String str : products) {
insertWord(str);
}
}
​
public void insertWord(String word) {
TrieNode node = root;
for (char c : word.toCharArray()) {
if (node.children[c - 'a'] == null) {
node.children[c - 'a'] = new TrieNode();
}
node = node.children[c - 'a'];
}
if (node.end != true) {
node.end = true;
node.str = word;
}
node.count++;
}
​
public List<List<String>> searchWord(String word) {
List<List<String>> result = new ArrayList<>();
for (int i = 1; i <= word.length(); i++) {
result.add(search(word.substring(0, i)));
}
return result;
}
​
private List<String> search(String pattern) {
List<String> result = new ArrayList<>();
TrieNode node = root;
for (char c : pattern.toCharArray()) {
if (node.children[c - 'a'] == null) {
return result;
}
node = node.children[c - 'a'];
}
​
Solution(node, result);
return result;
}
​
private void Solution(TrieNode root, List<String> result) {
if (root.end) {
for (int i = 0; i < root.count; i++) {
result.add(root.str);
if (result.size() == 3) {
return;
}
}
}
​
for (TrieNode node : root.children) {
if (node != null) {
Solution(node, result);
}
if (result.size() == 3) {
return;
}
}
​
}
}
​
public List<List<String>> suggestedProducts(String[] products, String searchWord) {
Trie trie = new Trie();
trie.insert(products);
return trie.searchWord(searchWord);
}
}