Python algorithm-Insert sort

Insert Sort

The core idea: in an ordered array, the position of the new number is found by comparing one to the previous number.

Example: An array has a number 21

Insert a 3,3<21, so the result is 3,21

then insert a 34,34>21, So the result is 3,21,34

then insert a 6,34>6, move forward,21>6, move forward,3<6, So the result is 3,6,21,34

question 1: How many times do you need to insert an array with n numbers to complete the insertion sort?

N-1 Times

Second, the core code:

def insert_sort (lista):

For I in Xrange (1,len (Lista)): # use i to control the subscript from the second number to the last number

j = i-1 # with J to control I front one number and i do compare

While J >= 0: #j <0 is a decision condition for exit loops

If LISTA[J] > lista[j+1]: # If the first number is greater than the last number, swap position

LISTA[J],LISTA[J+1] = Lista[j+1],lista[j]

J-= 1 #j minus one to compare to the previous value

Else

Break # As soon as there is a j>j+1, swap out this cycle can be

return lista

if __name__ = = ' __main__ ':

Print Insert_sort ([2,15,6,29,0])

Three, time complexity:

worst case comparison 1+2+ ... +n-2 + n-1 = N (n-1)/2 N^2/2

the best comparison situation is n-1 ..... exactly in the corresponding position

The average situation is [N (n-1)/2 + (n-1)]/2 = N^2/4

The final result is O (n^2)

Python algorithm-Insert sort