Python USES recursion to solve the all permutation number example

The first method: recursion

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

[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]

The second approach: the python standard library

import itertools
print list(itertools.permutations([1, 2, 3,4],3))

The source code is 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