Python uses recursion to solve all-ranked digital examples _python

The first method: recursion

Copy Code code as follows:

def perms (elements):
If Len (elements) <=1:
Yield elements
Else
For Perm in Perms (elements[1:]):
For I in range (len (elements)):
Yield perm[:i] + elements[0:1] + perm[i:]

For item in list (perms [1, 2, 3,4]):
Print Item

Results

Copy Code code as follows:

[1, 2, 3, 4]
[2, 1, 3, 4]
[2, 3, 1, 4]
[2, 3, 4, 1]
[1, 3, 2, 4]
[3, 1, 2, 4]
[3, 2, 1, 4]
[3, 2, 4, 1]
[1, 3, 4, 2]
[3, 1, 4, 2]
[3, 4, 1, 2]
[3, 4, 2, 1]
[1, 2, 4, 3]
[2, 1, 4, 3]
[2, 4, 1, 3]
[2, 4, 3, 1]
[1, 4, 2, 3]
[4, 1, 2, 3]
[4, 2, 1, 3]
[4, 2, 3, 1]
[1, 4, 3, 2]
[4, 1, 3, 2]
[4, 3, 1, 2]
[4, 3, 2, 1]

Second method: Python standard library

Copy Code code as follows:

Import Itertools
Print List (Itertools.permutations ([1, 2, 3,4],3))

The source code is as follows:

Copy Code code as follows:

#coding: Utf-8
Import Itertools
Print List (Itertools.permutations ([1, 2, 3,4],3))

def perms (elements):
If Len (elements) <=1:
Yield elements
Else
For Perm in Perms (elements[1:]):
For I in range (len (elements)):
Yield perm[:i] + elements[0:1] + perm[i:]

For item in list (perms [1, 2, 3,4]):
Print Item