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https://gitee.com/coder-xiaomo/leetcode-problemset
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92 lines
10 KiB
JSON
92 lines
10 KiB
JSON
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"questionId": "1182",
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"questionFrontendId": "550",
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"categoryTitle": "Database",
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"boundTopicId": 11038,
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"title": "Game Play Analysis IV",
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"titleSlug": "game-play-analysis-iv",
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"content": "<p>Table: <code>Activity</code></p>\n\n<pre>\n+--------------+---------+\n| Column Name | Type |\n+--------------+---------+\n| player_id | int |\n| device_id | int |\n| event_date | date |\n| games_played | int |\n+--------------+---------+\n(player_id, event_date) is the primary key (combination of columns with unique values) of this table.\nThis table shows the activity of players of some games.\nEach row is a record of a player who logged in and played a number of games (possibly 0) before logging out on someday using some device.\n</pre>\n\n<p> </p>\n\n<p>Write a solution to report the <strong>fraction</strong> of players that logged in again on the day after the day they first logged in, <strong>rounded to 2 decimal places</strong>. In other words, you need to determine the number of players who logged in on the day immediately following their initial login, and divide it by the number of total players.</p>\n\n<p>The result format is in the following example.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nActivity table:\n+-----------+-----------+------------+--------------+\n| player_id | device_id | event_date | games_played |\n+-----------+-----------+------------+--------------+\n| 1 | 2 | 2016-03-01 | 5 |\n| 1 | 2 | 2016-03-02 | 6 |\n| 2 | 3 | 2017-06-25 | 1 |\n| 3 | 1 | 2016-03-02 | 0 |\n| 3 | 4 | 2018-07-03 | 5 |\n+-----------+-----------+------------+--------------+\n<strong>Output:</strong> \n+-----------+\n| fraction |\n+-----------+\n| 0.33 |\n+-----------+\n<strong>Explanation:</strong> \nOnly the player with id 1 logged back in after the first day he had logged in so the answer is 1/3 = 0.33\n</pre>\n",
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"translatedTitle": "游戏玩法分析 IV",
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"translatedContent": "<p>Table: <code>Activity</code></p>\n\n<pre>\n+--------------+---------+\n| Column Name | Type |\n+--------------+---------+\n| player_id | int |\n| device_id | int |\n| event_date | date |\n| games_played | int |\n+--------------+---------+\n（player_id，event_date）是此表的主键（具有唯一值的列的组合）。\n这张表显示了某些游戏的玩家的活动情况。\n每一行是一个玩家的记录，他在某一天使用某个设备注销之前登录并玩了很多游戏（可能是 0）。\n</pre>\n\n<p>编写解决方案，报告在首次登录的第二天再次登录的玩家的 <strong>比率</strong>，<strong>四舍五入到小数点后两位</strong>。换句话说，你需要计算从首次登录后的第二天登录的玩家数量，并将其除以总玩家数。</p>\n\n<p>结果格式如下所示：</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>\nActivity table:\n+-----------+-----------+------------+--------------+\n| player_id | device_id | event_date | games_played |\n+-----------+-----------+------------+--------------+\n| 1 | 2 | 2016-03-01 | 5 |\n| 1 | 2 | 2016-03-02 | 6 |\n| 2 | 3 | 2017-06-25 | 1 |\n| 3 | 1 | 2016-03-02 | 0 |\n| 3 | 4 | 2018-07-03 | 5 |\n+-----------+-----------+------------+--------------+\n<strong>输出：</strong>\n+-----------+\n| fraction |\n+-----------+\n| 0.33 |\n+-----------+\n<strong>解释：</strong>\n只有 ID 为 1 的玩家在第一天登录后才重新登录，所以答案是 1/3 = 0.33\n</pre>\n",
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"sampleTestCase": "{\"headers\":{\"Activity\":[\"player_id\",\"device_id\",\"event_date\",\"games_played\"]},\"rows\":{\"Activity\":[[1,2,\"2016-03-01\",5],[1,2,\"2016-03-02\",6],[2,3,\"2017-06-25\",1],[3,1,\"2016-03-02\",0],[3,4,\"2018-07-03\",5]]}}",
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"Create table If Not Exists Activity (player_id int, device_id int, event_date date, games_played int)",
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"insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-01', '5')",
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"insert into Activity (player_id, device_id, event_date, games_played) values ('1', '2', '2016-03-02', '6')",
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"insert into Activity (player_id, device_id, event_date, games_played) values ('2', '3', '2017-06-25', '1')",
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"insert into Activity (player_id, device_id, event_date, games_played) values ('3', '1', '2016-03-02', '0')",
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"insert into Activity (player_id, device_id, event_date, games_played) values ('3', '4', '2018-07-03', '5')"
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