2019.12.9 LeetCode 从零单刷个人笔记整理（持续更新）
github：https://github.com/ChopinXBP/LeetCode-Babel
这是参加的第二场周赛。
本题很直接可以想到前缀树+DFS的思路，不过要注意一点神坑：测试用例的字典中可能包含重复的单词。应对这点，只需要在字典树结点中加上条件count用于统计相同单词的出现次数即可。
也可以不直接建树，先将字典单词排序，每次遍历前缀符合的单词，保留集合以供下一轮遍历。多次重复遍历即可得出结果。
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.
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。
import java.util.ArrayList;
import java.util.List;
/**
*
* 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.
* 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。
*
*/
public class SearchSuggestionsSystem {
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);
}
}
private void insertWord(String products){
TrieNode node = root;
for(char c : products.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 = products;
}
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);
}
}
#Coding一小时，Copying一秒钟。留个言点个赞呗，谢谢你#
