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'''
Author: SJ2050
Date: 2021-10-10 23:34:50
LastEditTime: 2021-10-16 23:23:48
Version: v0.0.1
Description: Solution for homework 2.
Copyright © 2021 SJ2050
'''
from itertools import permutations
def combination ( arr , n ) :
if n == 0 :
return [ [ ] ]
l = [ ]
for i in range ( 0 , len ( arr ) ) :
m = arr [ i ]
remLst = arr [ i + 1 : ]
for p in combination ( remLst , n - 1 ) :
l . append ( [ m ] + p )
return l
def heapPermutation ( arr , n , process ) :
if n == 1 :
process ( arr )
return
for i in range ( n ) :
heapPermutation ( arr , n - 1 , process )
if n & 1 :
arr [ 0 ] , arr [ n - 1 ] = arr [ n - 1 ] , arr [ 0 ]
else :
arr [ i ] , arr [ n - 1 ] = arr [ n - 1 ] , arr [ i ]
if __name__ == ' __main__ ' :
nums = [ 1 , 2 , 3 , 4 ]
m = 3
count = 0
combs = combination ( nums , m )
def printAndCount ( arr ) :
global count
count + = 1
for i in range ( len ( arr ) ) :
print ( f ' { arr [ i ] } ' , end = ' ' )
print ( ' ' )
print ( ' 3-digit nums: ' )
for c in combs :
heapPermutation ( c , m , printAndCount )
print ( ' - ' * 12 )
print ( f ' Total count: { count } ' )
# 从一列数中取出三个元素组成三位数的算法复杂度分析:
# 首先是组合分析， ，
# 接着， ， ，
# 故总的时间复杂度为: O((n^3)*(3!)) --> O(n^3)
