[Swift]LeetCode1268. 搜索推荐系统 | Search Suggestions System

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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"]
Example 2:

Input: products = ["havana"], searchWord = "havana"
Output: [["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]
Example 3:

Input: products = ["bags","baggage","banner","box","cloths"], searchWord = "bags"
Output: [["baggage","bags","banner"],["baggage","bags","banner"],["baggage","bags"],["bags"]]
Example 4:

Input: products = ["havana"], searchWord = "tatiana"
Output: [[],[],[],[],[],[],[]]

Constraints:

1 <= products.length <= 1000
1 <= Σ products[i].length <= 2 * 10^4
All characters of products[i] are lower-case English letters.
1 <= searchWord.length <= 1000
All characters of searchWord are lower-case English letters.

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。

 1 class Solution {
 2     func suggestedProducts(_ products: [String], _ searchWord: String) -> [[String]] {
 3         let products = products.sorted(by:<)
 4         var res:[[String]] = [[String]]()
 5         var sb:String = String()
 6         let cs:[Character] = Array(searchWord)
 7         var idx:Int = 0
 8         for c in cs
 9         {
10             sb.append(c)
11             let curSearch = sb
12             var cur:[String] = [String]()
13             if idx >= products.count
14             {
15                 res.append(cur)
16                 continue
17             }
18             while (idx < products.count && products[idx] < curSearch)
19             {
20                 idx += 1
21             }
22             var idxCopy:Int = idx
23             for _ in 0..<3
24             {
25                 if idxCopy >= products.count
26                 {
27                     break
28                 }
29                 if products[idxCopy].hasPrefix(curSearch)
30                 {
31                     cur.append(products[idxCopy])
32                     idxCopy += 1
33                 }
34                 else
35                 {
36                     break
37                 }
38             }
39             res.append(cur)
40         }
41         return res
42     }
43 }