[Function add]

1. Add leetcode solutions.

Author: Seanforfun

README.md

@@ -313,6 +313,12 @@
 
 [175. Combine Two Tables](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/175.%20Combine%20Two%20Tables.md)
 
+[176. Second Highest Salary](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/176.%20Second%20Highest%20Salary.md)
+
+[177. Nth Highest Salary](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/177.%20Nth%20Highest%20Salary.md)
+
+[178. Rank Scores](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/178.%20Rank%20Scores.md)
+
 ## Algorithm(4th_Edition)
 Reading notes of book Algorithm(4th Algorithm),ISBN: 9787115293800.
 All java realization codes are placed in different packages.

leetcode/176. Second Highest Salary.md

@@ -52,4 +52,18 @@ where
 not in(
     select max(Salary) from Employee
 )
+```
+
+### 二刷
+1. 通过子查询实现筛选出第二大的数。
+2. 通过IFNULL判断，如果不存在则填写NULL。
+3. 使用DISTINCT去重。
+
+```SQL
+# Write your MySQL query statement below
+SELECT IFNULL(
+(SELECT distinct Salary
+FROM Employee
+ORDER BY Salary desc limit 1,1)
+, null) as SecondHighestSalary ;
 ```

leetcode/177. Nth Highest Salary.md

@@ -42,4 +42,21 @@ set M = N - 1;
             M, 1
   );
 END
+```
+
+### 二刷
+1. offset关键字后的数据不能进行数学操作。这才是我们需要在外部对N - 1进行赋值的原因。
+```SQL
+CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
+BEGIN
+DECLARE M int;
+SET M = N - 1;
+  RETURN (
+      # Write your MySQL query statement below.
+      SELECT IFNULL(
+          (SELECT DISTINCT Salary
+          FROM Employee
+          ORDER BY Salary DESC LIMIT 1 OFFSET M), NULL) as Salary
+  );
+END
 ```

leetcode/178. Rank Scores.md

@@ -0,0 +1,46 @@
+## 178. Rank Scores
+
+### Question
+Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" between ranks.
+
+```
++----+-------+
+| Id | Score |
++----+-------+
+| 1  | 3.50  |
+| 2  | 3.65  |
+| 3  | 4.00  |
+| 4  | 3.85  |
+| 5  | 4.00  |
+| 6  | 3.65  |
++----+-------+
+```
+
+For example, given the above Scores table, your query should generate the following report (order by highest score):
+```
++-------+------+
+| Score | Rank |
++-------+------+
+| 4.00  | 1    |
+| 4.00  | 1    |
+| 3.85  | 2    |
+| 3.65  | 3    |
+| 3.65  | 3    |
+| 3.50  | 4    |
++-------+------+
+```
+
+### 二刷：
+1. 这道题的核心在于自己创建新的变量。
+2. 这道题的还牵扯到SQL的优先级问题，我们单独拎出来@pre <> (@pre := Score)进行分析。
+3. 实际上这行的执行顺序是从左往右，并没有将括号内的优先拉出来运算。
+```SQL
+# Write your MySQL query statement below
+SELECT s.Score as Score,
+(@i:=@i+ (@pre <> (@pre := Score))) as Rank
+FROM Scores s, (select @i:=0, @pre:=-1) AS init
+ORDER BY s.Score DESC;
+```
+
+### 引用
+1. [MySQL中变量的用法——LeetCode 178. Rank Scores](http://www.cnblogs.com/rever/p/7149995.html)