Four sorts: bubble, select, insert, count

Bubble sort

Selection sort

Insertion sort

Count sorting

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Bubble sort

Ideas ： Take ascending order , Compare two adjacent numbers , If the previous number is greater than the latter , Then two numbers are exchanged , And so on , After the trip, the maximum number is put to the last . The first 2 This operation is repeated for the remaining number except the last one , You can put the second largest number to the penultimate 2 The location of . Outer loop control hierarchy , The inner loop finds the maximum value and exchanges .

s=[8,3,2,9,1]
for i in range(len(s)-1):
	for j in range(len(s)-i-1):
		if(s[j]>s[j+1]):
			s[j],s[j+1]=s[j+1],s[j]
print(s)

Selection sort

Ideas ： Take ascending order , First, find the minimum number and its position in a group , Then exchange with the first number , Then in the remaining numbers except the first number , Then find out the minimum number and position , And exchange with the number in the second position , Sort a group of numbers by analogy .

s=[8,3,2,9,1]
for i in range(0,len(s)-1):
	min=s[i]
	p=i
	for j in range(i+1,len(s)):
		if min>s[j]:
			min=s[j]
			p=j
	s[p]=s[i]
	s[i]=min
print(s)

Insertion sort

Ideas ： Take ascending order , Will be the first i Elements and number i-1 Compare elements , Ruodi i One element is better than the other i-1 It's a small element , The first i-1 The first element is the same as i The price of elements changes positions , After exchange, the i Elements are compared with the previous one , Exchange positions when you are smaller than the front , until i>0.

s=[8,3,2,9,1]
for i in range(1,len(s)):
	for j in range(i,0,-1):
		if(s[j]<s[j-1]):
			t=s[j]
			s[j]=s[j-1]
			s[j-1]=t
print(s)

Count sorting

principle ： In a new array of minimum to maximum values , Save the number of elements of the original array , Then, according to the number of elements of the original array saved by the new array , Output the original array elements in turn .

a=[3,-1,7,2,5,0]
max=a[0]
min=a[0]
for i in range(1,len(a)):
	if max<a[i]:
		max=a[i]
	if min>a[i]:
		min=a[i]
l=max-min+1 #b Length of array 
b=[0]*l     # because b Used to save arrays a The number of each element of , therefore b The initial value of the assignment is 0
for i in a:
	b[i-min]+=1  # Traverse a, Find the number of elements 
print(b)
for i in range(l):
	for j in range(0,b[i]):  #b[i] Saved array a Number of elements 
		print(i+min,end=' ')  # The output array a The elements of