597. 好友申请 I ：总体通过率

思路：

1. 统计申请表中的不重复行数,记为表A

(SELECT COUNT(*)
FROM (SELECT DISTINCT sender_id, send_to_id
      FROM friend_request) A)

2. 统计接受表中的不重复行数，记为表B

(SELECT COUNT(*)
FROM (SELECT DISTINCT requester_id, accepter_id
      FROM request_accepted) B)

3. 表B结果/表A结果，IFNULL（，0）表示为空输出0

最终代码

SELECT ROUND(IFNULL(
(SELECT COUNT(*)
FROM (SELECT DISTINCT requester_id, accepter_id
      FROM request_accepted) B)
/
(SELECT COUNT(*)
FROM (SELECT DISTINCT sender_id, send_to_id
      FROM friend_request) A)
,0) ,2) AS accept_rate;

602. 好友申请 II ：谁有最多的好友

思路：

采用UNION ALL将表的requester_id和accepter_id列上下拼接，并创建临时表A

用COUNT()统计每个requester_id出现的次数

将A按照requester_id降序排列

取第一行。

SELECT A.requester_id AS id, COUNT(A.requester_id) AS num
FROM(
SELECT requester_id
FROM request_accepted
UNION ALL
SELECT accepter_id
FROM request_accepted) A 
GROUP BY A.requester_id
ORDER BY num DESC
LIMIT 1;

[SQL]597(表2行数/表1行数)+602(表的上下拼接)

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