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
这里有一个算法,它在不创建新的中间列表的情况下处理列表,类似于Ber在https://stackoverflow.com/a/108651/184528.
def permute(xs, low=0):
if low + 1 >= len(xs):
yield xs
else:
for p in permute(xs, low + 1):
yield p
for i in range(low + 1, len(xs)):
xs[low], xs[i] = xs[i], xs[low]
for p in permute(xs, low + 1):
yield p
xs[low], xs[i] = xs[i], xs[low]
for p in permute([1, 2, 3, 4]):
print p
您可以在这里亲自尝试代码:http://repl.it/J9v
其他回答
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']
def permuteArray (arr):
arraySize = len(arr)
permutedList = []
if arraySize == 1:
return [arr]
i = 0
for item in arr:
for elem in permuteArray(arr[:i] + arr[i + 1:]):
permutedList.append([item] + elem)
i = i + 1
return permutedList
我不打算在一个新的行中穷尽所有的可能性,以使它有点独特。
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
使用标准库中的itertools.permutations:
import itertools
list(itertools.permutations([1, 2, 3]))
从这里改编的是itertools.permutations如何实现的演示:
def permutations(elements):
if len(elements) <= 1:
yield elements
return
for perm in permutations(elements[1:]):
for i in range(len(elements)):
# nb elements[0:1] works in both string and list contexts
yield perm[:i] + elements[0:1] + perm[i:]
itertools.permutations文档中列出了两种替代方法
def permutations(iterable, r=None):
# permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
# permutations(range(3)) --> 012 021 102 120 201 210
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
if r > n:
return
indices = range(n)
cycles = range(n, n-r, -1)
yield tuple(pool[i] for i in indices[:r])
while n:
for i in reversed(range(r)):
cycles[i] -= 1
if cycles[i] == 0:
indices[i:] = indices[i+1:] + indices[i:i+1]
cycles[i] = n - i
else:
j = cycles[i]
indices[i], indices[-j] = indices[-j], indices[i]
yield tuple(pool[i] for i in indices[:r])
break
else:
return
另一个基于itertools.product:
def permutations(iterable, r=None):
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
for indices in product(range(n), repeat=r):
if len(set(indices)) == r:
yield tuple(pool[i] for i in indices)
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:]))