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
我的Python解决方案:
def permutes(input,offset):
if( len(input) == offset ):
return [''.join(input)]
result=[]
for i in range( offset, len(input) ):
input[offset], input[i] = input[i], input[offset]
result = result + permutes(input,offset+1)
input[offset], input[i] = input[i], input[offset]
return result
# input is a "string"
# return value is a list of strings
def permutations(input):
return permutes( list(input), 0 )
# Main Program
print( permutations("wxyz") )
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:]))
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。
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 permutate(l):
for i, x in enumerate(l):
for y in l[i + 1:]:
yield x, y
if __name__ == '__main__':
print(list(permutate(list('abcd'))))
print(list(permutate([1, 2, 3, 4])))
#[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
#[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]