[Swift]LeetCode1152. 用戶網站訪問行爲分析 | Analyze User Website Visit Pattern

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：爲敢（WeiGanTechnologies）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-alactdio-kp.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

You are given three arrays username, timestamp and website of the same length N where the ithtuple means that the user with name username[i] visited the website website[i] at time timestamp[i].

A 3-sequence is a list of not necessarily different websites of length 3 sorted in ascending order by the time of their visits.

Find the 3-sequence visited at least once by the largest number of users. If there is more than one solution, return the lexicographically minimum solution.

A 3-sequence X is lexicographically smaller than a 3-sequence Y if X[0] < Y[0] or X[0] == Y[0] and (X[1] < Y[1] or X[1] == Y[1] and X[2] < Y[2]). 

It is guaranteed that there is at least one user who visited at least 3 websites. No user visits two websites at the same time.

Example 1:

Input: username = ["joe","joe","joe","james","james","james","james","mary","mary","mary"], timestamp = [1,2,3,4,5,6,7,8,9,10], website = ["home","about","career","home","cart","maps","home","home","about","career"] Output: ["home","about","career"] Explanation: The tuples in this example are: ["joe", 1, "home"] ["joe", 2, "about"] ["joe", 3, "career"] ["james", 4, "home"] ["james", 5, "cart"] ["james", 6, "maps"] ["james", 7, "home"] ["mary", 8, "home"] ["mary", 9, "about"] ["mary", 10, "career"] The 3-sequence ("home", "about", "career") was visited at least once by 2 users. The 3-sequence ("home", "cart", "maps") was visited at least once by 1 user. The 3-sequence ("home", "cart", "home") was visited at least once by 1 user. The 3-sequence ("home", "maps", "home") was visited at least once by 1 user. The 3-sequence ("cart", "maps", "home") was visited at least once by 1 user.

Note:

1. 3 <= N = username.length = timestamp.length = website.length <= 50
2. 1 <= username[i].length <= 10
3. 0 <= timestamp[i] <= 10^9
4. 1 <= website[i].length <= 10
5. Both username[i] and website[i] contain only lowercase characters.

---

爲了評估某網站的用戶轉化率，咱們須要對用戶的訪問行爲進行分析，並創建用戶行爲模型。日誌文件中已經記錄了用戶名、訪問時間 以及 頁面路徑。

爲了方便分析，日誌文件中的 N 條記錄已經被解析成三個長度相同且長度都爲 N 的數組，分別是：用戶名 username，訪問時間 timestamp 和 頁面路徑 website。第 i 條記錄意味着用戶名是 username[i] 的用戶在 timestamp[i] 的時候訪問了路徑爲 website[i] 的頁面。

咱們須要找到用戶訪問網站時的 『共性行爲路徑』，也就是有最多的用戶都 至少按某種次序訪問過一次 的三個頁面路徑。須要注意的是，用戶 可能不是連續訪問 這三個路徑的。

『共性行爲路徑』是一個 長度爲 3 的頁面路徑列表，列表中的路徑 沒必要不一樣，而且按照訪問時間的前後升序排列。

若是有多個知足要求的答案，那麼就請返回按字典序排列最小的那個。（頁面路徑列表 X 按字典序小於 Y 的前提條件是：X[0] < Y[0] 或 X[0] == Y[0] 且 (X[1] < Y[1] 或 X[1] == Y[1] 且 X[2] < Y[2])）

題目保證一個用戶會至少訪問 3 個路徑一致的頁面，而且一個用戶不會在同一時間訪問兩個路徑不一樣的頁面。

示例：

輸入：username = ["joe","joe","joe","james","james","james","james","mary","mary","mary"], timestamp = [1,2,3,4,5,6,7,8,9,10], website = ["home","about","career","home","cart","maps","home","home","about","career"]
輸出：["home","about","career"]
解釋：
由示例輸入獲得的記錄以下：
["joe", 1, "home"]
["joe", 2, "about"]
["joe", 3, "career"]
["james", 4, "home"]
["james", 5, "cart"]
["james", 6, "maps"]
["james", 7, "home"]
["mary", 8, "home"]
["mary", 9, "about"]
["mary", 10, "career"]
有 2 個用戶至少訪問過一次 ("home", "about", "career")。
有 1 個用戶至少訪問過一次 ("home", "cart", "maps")。
有 1 個用戶至少訪問過一次 ("home", "cart", "home")。
有 1 個用戶至少訪問過一次 ("home", "maps", "home")。
有 1 個用戶至少訪問過一次 ("cart", "maps", "home")。

提示：

1. 3 <= N = username.length = timestamp.length = website.length <= 50
2. 1 <= username[i].length <= 10
3. 0 <= timestamp[i] <= 10^9
4. 1 <= website[i].length <= 10
5. username[i] 和 website[i] 都只含小寫字符

---

 

 1 class Solution {
 2     func mostVisitedPattern(_ username: [String], _ timestamp: [Int], _ website: [String]) -> [String] {
 3         let N:Int = username.count
 4         var webs:Set<String> = Set<String>(website)
 5         var v:[String:[Int:String]] = [String:[Int:String]]()
 6         for i in 0..<N
 7         {
 8             v[username[i],default:[Int:String]()][timestamp[i]] = website[i]
 9         }
10         var ans:Int = 0
11         var ret:[String] = [String]()
12         for a in website
13         {
14             for b in website
15             {
16                 for c in website
17                 {
18                     var num:Int = 0
19                     for (user, history) in v
20                     {
21                         var cnt:Int = 0
22                         for (time, web) in history
23                         {
24                             if cnt == 0 && web == a {cnt += 1}
25                             else if cnt == 1 && web == b {cnt += 1}
26                             else if cnt == 2 && web == c
27                             {
28                                 cnt += 1
29                                 break
30                             }   
31                         }
32                         if cnt == 3 {num += 1}      
33                     } 
34                     if num > ans
35                     {
36                         ans = num
37                         ret = [a, b, c]
38                     }
39                 }
40             }
41         }
42         return ret
43     }
44 }