Example:
c', 'b'] ['b', 'a', 'c'] ['b', 'c', 'a'] [ 'c', 'a', 'b'] ['c', 'b', 'a']
Method 1: Implemented using recursion
def permutation(li):
len_list = len(li)
if len_list == 1:
return li
result = []
for i in range(len_list):
res_list = li[:i] + li[i+1:]
s = li[i]
per_result = permutation(res_list)
if len(per_result) == 1:
(li[i:i + 1] + per_result)
else:
result += [[s] + j for j in per_result]
return result
Method 2: Utilizing python's own modules
import itertools def permutation(li): print(list((li)))
Additional extensions: python implementation of a full permutation of four numbers
First we use the regular practice of cyclic exchange completion.
lst = [1, 3, 5, 8]
for i in range(0, len(lst)):
lst[i], lst[0] = lst[0], lst[i]
for j in range(1, len(lst)):
lst[j], lst[1] = lst[1], lst[j]
for h in range(2, len(lst)):
print(lst)
lst[j], lst[1] = lst[1], lst[j]
lst[i], lst[0] = lst[0], lst[i]
If the list is longer, more elements, the above conventional method to achieve a more strenuous, the following we use recursive way to achieve.
def permutations(position):
if position == len(lst) - 1:
print(lst)
else:
for index in range(position, len(lst)):
lst[index], lst[position] = lst[position], lst[index]
permutations(position+1)
lst[index], lst[position] = lst[position], lst[index]
permutations(0)
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