1. Sum of three numbers

Source: LeetCode
Link: https://leetcode-cn.com/problems/3sum
Code: learn Wu Yanzu's solution

Give you an array num containing n integers. Judge whether there are three elements a, b and c in num, so that a + b + c = 0? Please find all triples with sum 0 and no repetition.

class Solution:
	def threeSum(self, nums: List[int]) -> List[List[int]]:
		n = len(nums) #Find list length
    	res = [] #Store in a place that meets the requirements
        if (not nums) or (n<3): #If the array is null or the array length is less than 3
            return [] #Returns []
        nums.sort() #Sort array
        if nums[0]>0: #Because the list is sorted well, when the first number is > 0, it cannot be greater than 0 after three numbers
            return []
        for i in range(n): #Traverse the sorted array
            if i>0 and nums[i]==nums[i-1]: #Prevent duplicate elements
                continue
            L=i+1 #Set left pointer
            R=n-1 #Set right pointer
            while(L<R):
                if(nums[i]+nums[L]+nums[R]==0): #If the sum of three is equal to 0
                    res.append([nums[i],nums[L],nums[R]) #Add three numbers to the res list
                    while(L<R and nums[L]==nums[L+1]): #Execute a loop to determine whether the left boundary repeats with the next position
                    	L=L+1
                    while(L<R and nums[R]==nums[R-1]): #Execute the loop to determine whether the right loop is repeated with the next position
                        R=R-1
                    #Move L and R to the next position to find a new solution
                    L = L+1
                    R = R-1
                elif(nums[i]+nums[L]+nums[R])>0):
                    #If the sum of the three numbers is greater than 0, the right value is too large. R should be smaller and move to the left
                    R = R-1
                else:
                    #If the sum of the three numbers is less than 0, the left value is too small. L should be a little larger and move to the right
                    L = L+1
        return res

2. Put apples

Source: niuke.com

Put m same apples on n same plates, and some plates are allowed to remain empty. How many different methods are there? (represented by K) 5, 1, 1 and 1, 5, 1 are the same division.

def f(m,n):
    if m<0 or n<0:
        return 0
    elif m==1 or n==1:
        return 1
    else:
    	#There are two situations for putting apples. One is that there are empty plates, and the other is that there are apples on each plate
    	#f(m,n-1) is the method for a plate to be empty
    	#f(m-n,n) is that there are apples on each plate
        return f(m,n-1)+f(m-n,n)
while True:
    try:
        m,n=map(int,input().split()) #m is the apple and n is the plate
        print(f(m,n))
    except:
        break

3. Recursion

Source: geek time - the beauty of data structure and algorithm
PS: This is a study note

The key of recursive code is to find the law of how to decompose a large problem into a small problem, write a recursive formula based on this, then refine the termination conditions, and finally translate the recursive formula and termination conditions into code.
There are three conditions for recursion:

1. The solution of a problem can be decomposed into the solutions of several subproblems
2. This problem is exactly the same as the sub problem after decomposition, except that the data scale is different
3. Recursive termination condition exists
   Example: climbing stairs
   How many ways to climb stairs are there for climbing one or two stairs, n steps?
   All walking methods can be divided into two categories according to the first step. The first is that the first step has taken one step, and the other is that the first step has taken two steps. Therefore, the walking method of N steps is equal to the walking method of n-1 steps after walking 1 step first, plus the walking method of n-2 steps after walking 2 steps first

f(n) = f(n-1)+f(n-2)

The termination condition is f(1)=1. If there are two steps, f(2)=f(1)+f(0), but there is no f(0), so the termination condition includes f(2)=2

4. Database learning

Source: look through your previous notes. The previous notes seem to look at which learning document you forgot before

Fuzzy query

select Listing from Table name where Listing like condition;
% Represents 0 or more arbitrary characters
_Represents any 1 character

Aggregate (grouping) function

count(Listing) Statistics
sum(Listing) Sum
avg(Listing) Average
min(Listing) Find the minimum value
max(Listing) Find the maximum value

Grouping query

select Listing/Aggregate function
from Table name
where condition
group by Listing
order by Column name 1/Aggregate function asc/desc;

having statement

After the grouping query results are obtained, the data is filtered again
select Listing/Aggregate function
from Table name
where condition
group by Listing
having condition
having Indicates that the grouped array is filtered and cannot be used alone. It must be combined with group by Use together

Sub (nested) query

It's in one SQL Statement contains another SQL sentence
 For example: query emp In the table, if the salary is greater than the average salary, the employee's number, name, position and salary
1,First, find the average wage
select avg(sal) from emp;
2,synthesis
select empno,ename,job,sal
from emp
where sal>(select avg(sal) from emp);

join query connection usage

inner join Internal connection to obtain the records of the field matching relationship in the two tables
left join Get all records in the left table, even if there is no corresponding matching record in the right table
right join Get all records in the right table, even if there is no corresponding matching record in the left table