Consecutive Numbers

Author: int32bit

README.md

@@ -53,6 +53,7 @@
 + [176 Second Highest Salary (子查询)](database/SecondHighestSalary)
 + [177 Nth Highest Salary(定义函数，变量，分支结构)](database/NthHighestSalary)
 + [178 Rank Scores (导出表、聚集)](database/RankScores)
++ [180 Consecutive Numbers(case语句)](database/ConsecutiveNumbers)
 + [181 Employees Earning More Than Their Managers(自连接)](database/EmployeesEarningMoreThanTheirManagers)
 + [182 Duplicate Emails(分组过滤)](database/DuplicateEmails)
 + [183 Customers Who Never Order(exists子句)](database/CustomersWhoNeverOrder)

database/ConsecutiveNumbers/README.md

@@ -0,0 +1,65 @@
+## Consecutive Numbers
+
+Write a SQL query to find all numbers that appear at least three times consecutively.
+```
++----+-----+
+| Id | Num |
++----+-----+
+| 1  |  1  |
+| 2  |  1  |
+| 3  |  1  |
+| 4  |  2  |
+| 5  |  1  |
+| 6  |  2  |
+| 7  |  2  |
++----+-----+
+```
+For example, given the above Logs table, 1 is the only number that appears consecutively for at least three times.
+
+## Solution 1
+
+使用自连接，然后一一和上一个比较，最后分组过滤即可
+```sql
+select log1.Num from Logs log1 join Logs log2 on log1.Id + 1 = log2.Id and log1.Num = log2.Num group by log1.Num having count(log1.Num) >= 2;
+```
+
+结果超时, 可能是分组导致的。
+
+## Solution 2
+
+还是使用自连接，不过为了避免group开销，直接使用三个表连接，分别指向Id, Id + 1, Id + 2，并分别让Num相等即可
+
+```sql
+select distinct a.Num from Logs a, Logs b,Logs c where a.Id=b.Id+1 and a.Num=b.Num and b.Id=c.Id+1 and b.Num=c.Num;
+```
+
+## Solution 3
+
+统计法。如果这一题使用c编程实现，逐行读取，这就容易了。
+
+可以使用case语句实现:
+```sql
+select Num from (
+SELECT
+        Num,
+        CASE
+            WHEN @prevNum = Num THEN @count := @count + 1
+            WHEN @prevNum := Num THEN @count := 1
+        END n
+FROM Logs, (select @prevNum := NULL) initPrevNum, (select @count := 1) initCount
+) as result where result.n >= 3;
+```
+注意在case后的结果别名为n，当不能直接在该语句中使用，必须使用另一个select语句查询。
+`when`是一个短路的，相当于`if`。`:=`用于赋值，`=`用于比较,必须在`from`语句中初始化`prevNum, count`的值
+
+## mysql case 语句
+
+相当于switch语句，when后面结条件，并且是短路的。例子:
+```sql
+select Num,
+	case
+		when Num % 2 = 0 then "Even"
+		when Num % 2 = 1 then "Odd"
+	end Numtype 
+from Logs;
+```

database/ConsecutiveNumbers/consecutiveNumbers.sql

@@ -1,11 +1,49 @@
 use leetcode;
-create table if not exists Logs(Id int, Num int);
-/*
+drop table if exists Logs;
+create table Logs(Id int, Num int);
 insert into Logs values(1,1);
 insert into Logs values(2,1);
 insert into Logs values(3,1);
 insert into Logs values(4,2);
 insert into Logs values(5,1);
 insert into Logs values(6,2);
 insert into Logs values(7,2);
-*/
+insert into Logs values(8,2);
+insert into Logs values(9,1);
+insert into Logs values(10,1);
+insert into Logs values(11,1);
+insert into Logs values(12,3);
+insert into Logs values(13,3);
+insert into Logs values(14,3);
+insert into Logs values(15,4);
+insert into Logs values(16,4);
+insert into Logs values(17,5);
+insert into Logs values(18,5);
+
+select "Method 1";
+select log1.Num from Logs log1 join Logs log2 on log1.Id + 1 = log2.Id and log1.Num = log2.Num group by log1.Num having count(log1.Num) >= 2;
+
+select "Method 2";
+select distinct a.Num from Logs a, Logs b,Logs c where a.Id=b.Id+1 and a.Num=b.Num and b.Id=c.Id+1 and b.Num=c.Num;
+
+select "Method 3";
+SELECT DISTINCT `Num` FROM (
+    SELECT
+        `Num`,
+        CASE
+            WHEN @prevNum = `Num` THEN @count := @count + 1
+            WHEN (@prevNum := `Num`) IS NOT NULL THEN @count := 1
+        END n
+    FROM `Logs`, (SELECT @prevNum := NULL) r, (SELECT @count := 1) c
+) a WHERE n >= 3;
+
+select "Method 4";
+SELECT DISTINCT `Num` FROM (
+    SELECT
+        `Num`,
+        CASE
+            WHEN @prevNum = `Num` THEN @count := @count + 1
+            WHEN (@prevNum := `Num`) IS NOT NULL THEN @count := 1
+        END n
+    FROM `Logs`
+) a WHERE n >= 3;

database/ConsecutiveNumbers/when.sql

@@ -0,0 +1,29 @@
+use leetcode;
+select Num,
+	case
+		when Num % 2 = 0 then "Even"
+		when Num % 2 = 1 then "Odd"
+	end Numtype 
+from Logs;
+
+select Num from (
+SELECT
+        Num,
+        CASE
+            WHEN @prevNum = Num THEN @count := @count + 1
+            WHEN @prevNum := Num THEN @count := 1
+        END n
+FROM Logs, (select @prevNum := NULL) initPrevNum, (select @count := 1) initCount
+) as result where result.n >= 3;
+/*
+SELECT * FROM (
+    SELECT
+        `Num`,
+        CASE
+            WHEN @prevNum = `Num` THEN @count := @count + 1
+            WHEN @prevNum := `Num` THEN @count := 1
+        END n
+    FROM `Logs`, (select @prevNum := NULL) r, (select @count := 1) s
+) a;
+
+*/