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
递归之美:
>>> import copy
>>> def perm(prefix,rest):
... for e in rest:
... new_rest=copy.copy(rest)
... new_prefix=copy.copy(prefix)
... new_prefix.append(e)
... new_rest.remove(e)
... if len(new_rest) == 0:
... print new_prefix + new_rest
... continue
... perm(new_prefix,new_rest)
...
>>> perm([],['a','b','c','d'])
['a', 'b', 'c', 'd']
['a', 'b', 'd', 'c']
['a', 'c', 'b', 'd']
['a', 'c', 'd', 'b']
['a', 'd', 'b', 'c']
['a', 'd', 'c', 'b']
['b', 'a', 'c', 'd']
['b', 'a', 'd', 'c']
['b', 'c', 'a', 'd']
['b', 'c', 'd', 'a']
['b', 'd', 'a', 'c']
['b', 'd', 'c', 'a']
['c', 'a', 'b', 'd']
['c', 'a', 'd', 'b']
['c', 'b', 'a', 'd']
['c', 'b', 'd', 'a']
['c', 'd', 'a', 'b']
['c', 'd', 'b', 'a']
['d', 'a', 'b', 'c']
['d', 'a', 'c', 'b']
['d', 'b', 'a', 'c']
['d', 'b', 'c', 'a']
['d', 'c', 'a', 'b']
['d', 'c', 'b', 'a']
其他回答
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
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]]
用递归求解,遍历元素,取第i个元素,然后问自己:“其余项目的排列是什么”,直到没有更多的元素。
我在这里解释了解决方案:https://www.youtube.com/watch?v=_7GE7psS2b4
class Solution:
def permute(self,nums:List[int])->List[List[int]]:
res=[]
def dfs(nums,path):
if len(nums)==0:
res.append(path)
for i in range(len(nums)):
dfs(nums[:i]+nums[i+1:],path+[nums[i]])
dfs(nums,[])
return res
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。
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