如何生成列表的所有排列?例如:
permutations([])
[]
permutations([1])
[1]
permutations([1, 2])
[1, 2]
[2, 1]
permutations([1, 2, 3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
如何生成列表的所有排列?例如:
permutations([])
[]
permutations([1])
[1]
permutations([1, 2])
[1, 2]
[2, 1]
permutations([1, 2, 3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
当前回答
我看到在这些递归函数中进行了很多迭代,而不是纯粹的递归。。。
所以对于那些连一个循环都不能遵守的人来说,这里有一个粗略的、完全不必要的完全递归的解决方案
def all_insert(x, e, i=0):
return [x[0:i]+[e]+x[i:]] + all_insert(x,e,i+1) if i<len(x)+1 else []
def for_each(X, e):
return all_insert(X[0], e) + for_each(X[1:],e) if X else []
def permute(x):
return [x] if len(x) < 2 else for_each( permute(x[1:]) , x[0])
perms = permute([1,2,3])
其他回答
def pzip(c, seq):
result = []
for item in seq:
for i in range(len(item)+1):
result.append(item[i:]+c+item[:i])
return result
def perm(line):
seq = [c for c in line]
if len(seq) <=1 :
return seq
else:
return pzip(seq[0], perm(seq[1:]))
功能性风格
def addperm(x,l):
return [ l[0:i] + [x] + l[i:] for i in range(len(l)+1) ]
def perm(l):
if len(l) == 0:
return [[]]
return [x for y in perm(l[1:]) for x in addperm(l[0],y) ]
print perm([ i for i in range(3)])
结果:
[[0, 1, 2], [1, 0, 2], [1, 2, 0], [0, 2, 1], [2, 0, 1], [2, 1, 0]]
这是初始排序后生成排列的渐近最优方式O(n*n!)。
有n个!最多进行一次置换,且具有下一次置换(..),以O(n)时间复杂度运行
在3个步骤中,
找到最大的j,使a[j]可以增加以最小可行量增加a[j]找到扩展新a[0..j]的字典最少方法
'''
Lexicographic permutation generation
consider example array state of [1,5,6,4,3,2] for sorted [1,2,3,4,5,6]
after 56432(treat as number) ->nothing larger than 6432(using 6,4,3,2) beginning with 5
so 6 is next larger and 2345(least using numbers other than 6)
so [1, 6,2,3,4,5]
'''
def hasNextPermutation(array, len):
' Base Condition '
if(len ==1):
return False
'''
Set j = last-2 and find first j such that a[j] < a[j+1]
If no such j(j==-1) then we have visited all permutations
after this step a[j+1]>=..>=a[len-1] and a[j]<a[j+1]
a[j]=5 or j=1, 6>5>4>3>2
'''
j = len -2
while (j >= 0 and array[j] >= array[j + 1]):
j= j-1
if(j==-1):
return False
# print(f"After step 2 for j {j} {array}")
'''
decrease l (from n-1 to j) repeatedly until a[j]<a[l]
Then swap a[j], a[l]
a[l] is the smallest element > a[j] that can follow a[l]...a[j-1] in permutation
before swap we have a[j+1]>=..>=a[l-1]>=a[l]>a[j]>=a[l+1]>=..>=a[len-1]
after swap -> a[j+1]>=..>=a[l-1]>=a[j]>a[l]>=a[l+1]>=..>=a[len-1]
a[l]=6 or l=2, j=1 just before swap [1, 5, 6, 4, 3, 2]
after swap [1, 6, 5, 4, 3, 2] a[l]=5, a[j]=6
'''
l = len -1
while(array[j] >= array[l]):
l = l-1
# print(f"After step 3 for l={l}, j={j} before swap {array}")
array[j], array[l] = array[l], array[j]
# print(f"After step 3 for l={l} j={j} after swap {array}")
'''
Reverse a[j+1...len-1](both inclusive)
after reversing [1, 6, 2, 3, 4, 5]
'''
array[j+1:len] = reversed(array[j+1:len])
# print(f"After step 4 reversing {array}")
return True
array = [1,2,4,4,5]
array.sort()
len = len(array)
count =1
print(array)
'''
The algorithm visits every permutation in lexicographic order
generating one by one
'''
while(hasNextPermutation(array, len)):
print(array)
count = count +1
# The number of permutations will be n! if no duplicates are present, else less than that
# [1,4,3,3,2] -> 5!/2!=60
print(f"Number of permutations: {count}")
在我看来,一个很明显的方式可能是:
def permutList(l):
if not l:
return [[]]
res = []
for e in l:
temp = l[:]
temp.remove(e)
res.extend([[e] + r for r in permutList(temp)])
return res
如果不想使用内置方法,例如:
import itertools
list(itertools.permutations([1, 2, 3]))
你可以自己实现permute函数
from collections.abc import Iterable
def permute(iterable: Iterable[str]) -> set[str]:
perms = set()
if len(iterable) == 1:
return {*iterable}
for index, char in enumerate(iterable):
perms.update([char + perm for perm in permute(iterable[:index] + iterable[index + 1:])])
return perms
if __name__ == '__main__':
print(permute('abc'))
# {'bca', 'abc', 'cab', 'acb', 'cba', 'bac'}
print(permute(['1', '2', '3']))
# {'123', '312', '132', '321', '213', '231'}