Leetcode | 刷题日记(1)

深藏阁楼爱情的钟 2022-07-18 01:52 106阅读 0赞

# 本文记录个人刷题记录 #

推荐两个刷题网站：

地址：https://leetcode.com/

另外一个地址：http://www.lintcode.com/

1.Write a SQL query to get the second highest salary from the `Employee` table.

题目地址：https://leetcode.com/problems/second-highest-salary/

+----+--------+
    | Id | Salary |
    +----+--------+
    | 1  | 100    |
    | 2  | 200    |
    | 3  | 300    |
    +----+--------+

For example, given the above Employee table, the second highest salary is  `200` . If there is no second highest salary, then the query should return  `null` .

解:

先算出所有员工中最高的工资，然后在从非最高工资中的记录，取最大的一个，就是第二高工资了

select max(Salary) as 'SecondHighestSalary'
    from Employee where Salary<(select max(Salary) from Employee)

2. Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

地址:https://leetcode.com/problems/two-sum/

Example:

Given nums = [2, 7, 11, 15], target = 9,
    
    Because nums[0] + nums[1] = 2 + 7 = 9,
    return [0, 1].

解：

两层遍历，外层遍历是负责遍历每个元素，内层遍历是用于把当前 外层循环去取出的数与除它之外的元素相加，看是否等于目标数值，如果等于，就返回当前两个元素的索引，如果都没有命中，就返回空数组

public class Solution {
        public int[] twoSum(int[] nums, int target) {
             for(int i=0 ;i<nums.length;i++){
               for(int j=i+1;j<nums.length;j++)
                   if (nums[i] + nums[j] == target) {
                       return new int[]{
                           i,j
                       };
                   }
           }
          return new int[]{};
        }
    }

3. Combine Two Tables

Table: `Person`

Table:  `Address`

Write a SQL query for a report that provides the following information for each person in the Person table, regardless if there is an address for each of those people:

FirstName, LastName, City, State

解:

两个表通过PersonId关联查询即可，但是因为需要有所有Person表记录，所以这里要用leftjoin

SELECT pes.firstName as 'FirstName',
    pes.LastName as 'LastName', 
    adr.City as 'City', 
    adr.State as 'State'
    FROM Person pes
    LEFT JOIN Address adr
    on pes.personid = adr.personid;