通常,permutations 算法可能看起来有点像这样(Python):
def permutations(elements):
if elements:
for i, current in enumerate(elements):
front, back = elements[:i], elements[i+1:]
for perm in permutations(front + back):
yield [current] + perm
else:
yield []
您迭代列表,将每个元素作为第一个元素,并将它们与其余元素的所有排列组合。您可以轻松地对其进行修改,以便 elements 实际上是元素列表,而不是仅使用 current 元素,您可以从该列表中弹出第一个元素并将其余元素插入到递归调用中:
def ordered_permutations(elements):
if elements:
for i, current in enumerate(elements):
front, back = elements[:i], elements[i+1:]
first, rest = current[0], current[1:]
for perm in ordered_permutations(front + ([rest] if rest else []) + back):
yield [first] + perm
else:
yield []
ordered_permutations([['A', 'B'], ['C', 'D'], ['E'], ['F']]) 的结果:
['A', 'B', 'C', 'D', 'E', 'F']
['A', 'B', 'C', 'D', 'F', 'E']
['A', 'B', 'C', 'E', 'D', 'F']
[ ... some 173 more ... ]
['F', 'E', 'A', 'C', 'D', 'B']
['F', 'E', 'C', 'A', 'B', 'D']
['F', 'E', 'C', 'A', 'D', 'B']
['F', 'E', 'C', 'D', 'A', 'B']
但请注意,这将在每个递归调用中创建大量中间列表。相反,您可以使用堆栈,将第一个元素从堆栈中弹出并在递归调用之后将其放回。
def ordered_permutations_stack(elements):
if any(elements):
for current in elements:
if current:
first = current.pop()
for perm in ordered_permutations_stack(elements):
yield [first] + perm
current.append(first)
else:
yield []
代码也可能更容易掌握。在这种情况下,您必须保留子列表,即将其称为ordered_permutations_stack([['B', 'A'], ['D', 'C'], ['E'], ['F']])