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
此解决方案实现了一个生成器,以避免在内存中保留所有排列:
def permutations (orig_list):
if not isinstance(orig_list, list):
orig_list = list(orig_list)
yield orig_list
if len(orig_list) == 1:
return
for n in sorted(orig_list):
new_list = orig_list[:]
pos = new_list.index(n)
del(new_list[pos])
new_list.insert(0, n)
for resto in permutations(new_list[1:]):
if new_list[:1] + resto <> orig_list:
yield new_list[:1] + resto
其他回答
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]]
以下代码是给定列表的就地排列,作为生成器实现。由于它只返回对列表的引用,因此不应在生成器外部修改列表。该解决方案是非递归的,因此使用了低内存。还可以很好地处理输入列表中元素的多个副本。
def permute_in_place(a):
a.sort()
yield list(a)
if len(a) <= 1:
return
first = 0
last = len(a)
while 1:
i = last - 1
while 1:
i = i - 1
if a[i] < a[i+1]:
j = last - 1
while not (a[i] < a[j]):
j = j - 1
a[i], a[j] = a[j], a[i] # swap the values
r = a[i+1:last]
r.reverse()
a[i+1:last] = r
yield list(a)
break
if i == first:
a.reverse()
return
if __name__ == '__main__':
for n in range(5):
for a in permute_in_place(range(1, n+1)):
print a
print
for a in permute_in_place([0, 0, 1, 1, 1]):
print a
print
对于Python 2.6及以上版本:
import itertools
itertools.permutations([1, 2, 3])
这将作为生成器返回。使用列表(排列(xs))作为列表返回。
如果用户希望在列表中保留所有排列,可以使用以下代码:
def get_permutations(nums, p_list=[], temp_items=[]):
if not nums:
return
elif len(nums) == 1:
new_items = temp_items+[nums[0]]
p_list.append(new_items)
return
else:
for i in range(len(nums)):
temp_nums = nums[:i]+nums[i+1:]
new_temp_items = temp_items + [nums[i]]
get_permutations(temp_nums, p_list, new_temp_items)
nums = [1,2,3]
p_list = []
get_permutations(nums, p_list)
from __future__ import print_function
def perm(n):
p = []
for i in range(0,n+1):
p.append(i)
while True:
for i in range(1,n+1):
print(p[i], end=' ')
print("")
i = n - 1
found = 0
while (not found and i>0):
if p[i]<p[i+1]:
found = 1
else:
i = i - 1
k = n
while p[i]>p[k]:
k = k - 1
aux = p[i]
p[i] = p[k]
p[k] = aux
for j in range(1,(n-i)/2+1):
aux = p[i+j]
p[i+j] = p[n-j+1]
p[n-j+1] = aux
if not found:
break
perm(5)