[Swift]LeetCode1152. 用户网站访问行为分析 | Analyze User Website Visit Pattern

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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.

---

为了评估某网站的用户转化率，我们需要对用户的访问行为进行分析，并建立用户行为模型。日志文件中已经记录了用户名、访问时间 以及 页面路径。

为了方便分析，日志文件中的 N 条记录已经被解析成三个长度相同且长度都为 N 的数组，分别是：用户名 username，访问时间 timestamp 和 页面路径 website。第 i 条记录意味着用户名是 username[i] 的用户在 timestamp[i] 的时候访问了路径为 website[i] 的页面。

我们需要找到用户访问网站时的 『共性行为路径』，也就是有最多的用户都 至少按某种次序访问过一次 的三个页面路径。需要注意的是，用户 可能不是连续访问 这三个路径的。

『共性行为路径』是一个 长度为 3 的页面路径列表，列表中的路径 不必不同，并且按照访问时间的先后升序排列。

如果有多个满足要求的答案，那么就请返回按字典序排列最小的那个。（页面路径列表 X 按字典序小于 Y 的前提条件是：X[0] < Y[0] 或 X[0] == Y[0] 且 (X[1] < Y[1] 或 X[1] == Y[1] 且 X[2] < Y[2])）

题目保证一个用户会至少访问 3 个路径一致的页面，并且一个用户不会在同一时间访问两个路径不同的页面。

示例：

输入：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"]
输出：["home","about","career"]
解释：
由示例输入得到的记录如下：
["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 个用户至少访问过一次 ("home", "about", "career")。
有 1 个用户至少访问过一次 ("home", "cart", "maps")。
有 1 个用户至少访问过一次 ("home", "cart", "home")。
有 1 个用户至少访问过一次 ("home", "maps", "home")。
有 1 个用户至少访问过一次 ("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] 都只含小写字符

---

class Solution {
    func mostVisitedPattern(_ username: [String], _ timestamp: [Int], _ website: [String]) -> [String] {
        let N:Int = username.count
        var webs:Set<String> = Set<String>(website)
        var v:[String:[Int:String]] = [String:[Int:String]]()
        for i in 0..<N
        {
            v[username[i],default:[Int:String]()][timestamp[i]] = website[i]
        }
        var ans:Int = 0
        var ret:[String] = [String]()
        for a in website
        {
            for b in website
            {
                for c in website
                {
                    var num:Int = 0
                    for (user, history) in v
                    {
                        var cnt:Int = 0
                        for (time, web) in history
                        {
                            if cnt == 0 && web == a {cnt += 1}
                            else if cnt == 1 && web == b {cnt += 1}
                            else if cnt == 2 && web == c
                            {
                                cnt += 1
                                break
                            }   
                        }
                        if cnt == 3 {num += 1}      
                    } 
                    if num > ans
                    {
                        ans = num
                        ret = [a, b, c]
                    }
                }
            }
        }
        return ret
    }
}