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1. Add leetcode solution.
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‎leetcode/181. Employees Earning More Than Their Managers.md‎

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## 181. Employees Earning More Than Their Managers
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### Question
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The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
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+----+-------+--------+-----------+
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| Id | Name | Salary | ManagerId |
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+----+-------+--------+-----------+
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| 1 | Joe | 70000 | 3 |
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| 2 | Henry | 80000 | 4 |
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| 3 | Sam | 60000 | NULL |
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| 4 | Max | 90000 | NULL |
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+----+-------+--------+-----------+
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Given the Employee table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
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+----------+
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| Employee |
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+----------+
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| Joe |
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+----------+
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### Thinking:
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* Method: 左外连接，速度比where要快。
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```SQL
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# Write your MySQL query statement below
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select
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a.Name as Employee
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from Employee a
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left join
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Employee b
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on
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a.ManagerID = b.Id
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where
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a.Salary > b.Salary
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```
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* Method 2:通过查询确定判断条件，这种方法比较慢，经过两次查询，使用了两个session。
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```Java
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# Write your MySQL query statement below
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select
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name as Employee
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from
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Employee a
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where
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a.Salary > (
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select
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b.Salary
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from
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Employee b
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where
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a.ManagerId = b.Id
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);
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```

‎leetcode/182. Duplicate Emails.md‎

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## 181. Employees Earning More Than Their Managers
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### Question
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The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
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+----+-------+--------+-----------+
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| Id | Name | Salary | ManagerId |
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+----+-------+--------+-----------+
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| 1 | Joe | 70000 | 3 |
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| 2 | Henry | 80000 | 4 |
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| 3 | Sam | 60000 | NULL |
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| 4 | Max | 90000 | NULL |
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+----+-------+--------+-----------+
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Given the Employee table, a## 182. Duplicate Emails
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### Question
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Write a SQL query to find all duplicate emails in a table named Person.
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+----+---------+
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| Id | Email |
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+----+---------+
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| 1 | a@b.com |
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| 2 | c@d.com |
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| 3 | a@b.com |
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+----+---------+
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For example, your query should return the following for the above table:
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+---------+
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| Email |
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+---------+
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| a@b.com |
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+---------+
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Note: All emails are in lowercase.
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### Thinking:
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* Method
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```SQL
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# Write your MySQL query statement below
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select
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Email
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from
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Person
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group by
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Email
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having
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count(*) > 1
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```write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.
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+----------+
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| Employee |
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+----------+
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| Joe |
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+----------+
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### Thinking:
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* Method: 左外连接，速度比where要快。
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```SQL
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# Write your MySQL query statement below
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select
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a.Name as Employee
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from Employee a
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left join
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Employee b
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on
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a.ManagerID = b.Id
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where
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a.Salary > b.Salary
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```
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* Method 2:通过查询确定判断条件，这种方法比较慢，经过两次查询，使用了两个session。
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```Java
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# Write your MySQL query statement below
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select
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name as Employee
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from
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Employee a
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where
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a.Salary > (
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select
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b.Salary
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from
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Employee b
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where
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a.ManagerId = b.Id
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);
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```

‎leetcode/183. Customers Who Never Order.md‎

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## 183. Customers Who Never Order
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### Question
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Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.
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Table: Customers.
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+----+-------+
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| Id | Name |
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+----+-------+
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| 1 | Joe |
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| 2 | Henry |
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| 3 | Sam |
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| 4 | Max |
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+----+-------+
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Table: Orders.
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+----+------------+
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| Id | CustomerId |
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+----+------------+
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| 1 | 3 |
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| 2 | 1 |
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+----+------------+
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Using the above tables as example, return the following:
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+-----------+
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| Customers |
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+-----------+
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| Henry |
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| Max |
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+-----------+
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### Thinking:
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* Method
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```SQL
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# Write your MySQL query statement below
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select
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Name as Customers
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from
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Customers
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where
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Id
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not in(
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select CustomerId from Orders
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);
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```

‎leetcode/190. Reverse Bits.md‎

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## 190. Reverse Bits
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### Question
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Reverse bits of a given 32 bits unsigned integer.
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```
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Example:
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Input: 43261596
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Output: 964176192
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Explanation: 43261596 represented in binary as 00000010100101000001111010011100,
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return 964176192 represented in binary as 00111001011110000010100101000000.
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```
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Follow up:
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If this function is called many times, how would you optimize it?
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### Thinking:
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* Method 1: 首尾取出元素，并交换位置存入一个新的int型数据中。
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```Java
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public class Solution {
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// you need treat n as an unsigned value
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public int reverseBits(int n) {
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int low = 1;
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int high = 1 << 31;
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int result = 0;
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for(int i = 16; i >= 1; i--){
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int right = n & low;
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int left = n & high;
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left >>= i * 2 - 1;
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result += left;
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right <<= i * 2 - 1;
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result += right;
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low <<= 1;
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right >>= 1;
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}
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return result;
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}
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}
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```

‎leetcode/191. Number of 1 Bits.md‎

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## 191. Number of 1 Bits
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### Question
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Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).
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```
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Example 1:
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Input: 11
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Output: 3
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Explanation: Integer 11 has binary representation 00000000000000000000000000001011
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Example 2:
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Input: 128
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Output: 1
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Explanation: Integer 128 has binary representation 00000000000000000000000010000000
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```
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### Thinking:
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* Method：
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```Java
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public class Solution {
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// you need to treat n as an unsigned value
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public int hammingWeight(int n) {
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int count = 0;
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for(int i = 0; i <= 31; i++){
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count = ((n & 1) == 1)? count + 1: count;
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n >>>= 1;
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}
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return count;
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}
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}
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```

‎leetcode/200. Number of Islands.md‎

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## 200. Number of Islands
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### Question
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Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
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```
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Example 1:
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Input:
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11110
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11010
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11000
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00000
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Output: 1
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Example 2:
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Input:
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11000
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11000
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00100
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00011
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Output: 3
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```
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### Thinking:
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* Method：DFS AC
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```Java
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class Solution {
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private int[][] dir = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
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public int numIslands(char[][] grid) {
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if(grid == null || grid.length == 0) return 0;
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int height = grid.length;
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int width = grid[0].length;
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boolean[][] used = new boolean[height][width];
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int count = 0;
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LinkedList<Integer> q = new LinkedList<>();
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for(int i = 0; i < height; i++){
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for(int j = 0; j < width; j++){
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if(grid[i][j] == '1' && !used[i][j]){
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count ++;
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bfs(grid, used, i, j);
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}
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}
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}
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return count;
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}
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private void bfs(char[][] grid, boolean[][] used, int row, int col){
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if(!(row >= 0 && row < grid.length && col >= 0 && col < grid[0].length && grid[row][col] == '1' && !used[row][col]))
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return;
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used[row][col] = true;
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for(int i = 0; i < 4; i++){
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bfs(grid, used, row + dir[i][0], col + dir[i][1]);
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}
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}
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}
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```
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* Method 2: BFS TLE
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```Java
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class Solution {
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public int numIslands(char[][] grid) {
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int count = 0;
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int[][] dir = new int[][]{{1,0}, {0,1}, {-1, 0}, {0, -1}};
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if(grid == null || grid.length == 0) return 0;
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int height = grid.length;
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int width = grid[0].length;
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boolean[][] used = new boolean[height][width];
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LinkedList<Integer> queue = new LinkedList<>();
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for(int i = 0; i < height; i++){
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for(int j = 0; j < width; j++){
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if(grid[i][j] == '0' || used[i][j]) continue;
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else{
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count++;
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queue.add(i * width + j);
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while(!queue.isEmpty()){
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Integer cnt = queue.poll();
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int row = cnt / width;
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int col = cnt % width;
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used[row][col] = true;
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for(int p = 0; p < 4; p++){
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int tempRow = row, tempCol = col;
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tempRow += dir[p][0];
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tempCol += dir[p][1];
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if(tempRow >= 0 && tempRow < height && tempCol >= 0 && tempCol < width && !used[tempRow][tempCol] && grid[tempRow][tempCol] == '1')
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queue.add(tempRow * width + tempCol);
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}
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}
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}
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}
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}
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return count;
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}
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}
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```

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