zoukankan      html  css  js  c++  java
  • LeetCode 1152. Analyze User Website Visit Pattern

    原题链接在这里:https://leetcode.com/problems/analyze-user-website-visit-pattern/

    题目:

    We are given some website visits: the user with name username[i] visited the website website[i] at time timestamp[i].

    3-sequence is a list of websites of length 3 sorted in ascending order by the time of their visits.  (The websites in a 3-sequence are not necessarily distinct.)

    Find the 3-sequence visited by the largest number of users. If there is more than one solution, return the lexicographically smallest such 3-sequence.

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

    题解:

    Try to get the website sequence visted by most different users. sequence here means chronological order.

    First sort the data by timestamp.

    And put user and corresponding websites into HashMap<String, List<String>>.

    For each user, find all its combination of 3 websites.

    And each sequence within this combination, check its count of different user. If count is higher or lexicographically smaller, update the max sequence.

    The max sequence is the result.

    Time Complexity: O(n ^ 3). n = username.length. sort takes O(nlogn). getCom takes O(n ^ 3).

    Space: O(n ^ 3).

    AC Java:

     1 class Solution {
     2     public List<String> mostVisitedPattern(String[] username, int[] timestamp, String[] website) {
     3         int n = username.length;
     4         List<Pair> datas = new ArrayList<>();
     5         for(int i = 0; i < n; i++){
     6             datas.add(new Pair(username[i], timestamp[i], website[i]));
     7         }
     8         
     9         Collections.sort(datas, (a, b) -> a.time - b.time);
    10         HashMap<String, List<String>> userToWebs = new HashMap<>();
    11         for(Pair data : datas){
    12             userToWebs.putIfAbsent(data.user, new ArrayList<String>());
    13             userToWebs.get(data.user).add(data.web);
    14         }
    15         
    16         HashMap<String, Integer> seqToCount = new HashMap<>();
    17         
    18         int maxCount = 0;
    19         String maxSeq = "";
    20         for(Map.Entry<String, List<String>> entry : userToWebs.entrySet()){
    21             Set<String> seqCom = getCom(entry.getValue());
    22             for(String seq : seqCom){
    23                 seqToCount.put(seq, seqToCount.getOrDefault(seq, 0) + 1);
    24                 if(seqToCount.get(seq) > maxCount){
    25                     maxCount = seqToCount.get(seq);
    26                     maxSeq = seq;
    27                 }else if(seqToCount.get(seq) == maxCount && seq.compareTo(maxSeq) < 0){
    28                     maxSeq = seq;
    29                 }    
    30             }
    31         }
    32         
    33         List<String> res = new ArrayList<>();
    34         String [] webs = maxSeq.split(",");
    35         for(String w : webs){
    36             res.add(w);
    37         }
    38         
    39         return res;
    40     }
    41     
    42     private HashSet<String> getCom(List<String> webs){
    43         HashSet<String> res = new HashSet<>();
    44         int n = webs.size();
    45         for(int i = 0; i < n - 2; i++){
    46             for(int j = i + 1; j < n - 1; j++){
    47                 for(int k = j + 1; k < n; k++){
    48                     res.add(webs.get(i) + "," + webs.get(j) + "," + webs.get(k));
    49                 }
    50             }
    51         }
    52         
    53         return res;
    54     }
    55 }
    56 
    57 class Pair{
    58     String user;
    59     int time;
    60     String web;
    61     public Pair(String user, int time, String web){
    62         this.user = user;
    63         this.time = time;
    64         this.web = web;
    65     }
    66 }
  • 相关阅读:
    数学图形(1.25)cassini曲线
    数学图形(1.24)巴斯加线与蚶线
    数学图形(1.23)太极线
    webpack打包多个入口文件
    cnpm与npm的区别
    入门 Webpack,看这篇就够了
    protocol error, got 'n' as reply type byte + redis如何后台启动
    PHP执行系统外部命令函数:exec()、passthru()、system()、shell_exec()
    CentOS7.0+Zend Guard Loader for PHP 5.6环境搭建
    通过shell脚本进行数据库操作
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/12356343.html
Copyright © 2011-2022 走看看