考虑使用以下函数来置换列表中的数字:
def permute(numbers, N=0):
# base case
if N == len(numbers):
print numbers
return
for i in range(len(numbers)-N):
# swapping relevant elements
element=numbers.pop(N+i)
numbers.insert(N,element)
# recursive call
permute(numbers, N+1)
# swapping back relevant elements when backtracking
element=numbers.pop(N)
numbers.insert(N+i,element)
numbers=[1,2,3]
permute(numbers)
为什么当我运行上面的代码时,它产生正确的输出:
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
但是当我试图将结果累积到列表时:
def permute(numbers, permutations, N=0):
# base case
if N == len(numbers):
print numbers
permutations.append(numbers)
return
for i in range(len(numbers)-N):
# swapping relevant elements
element=numbers.pop(N+i)
numbers.insert(N,element)
# recursive call
permute(numbers, permutations, N+1)
# swapping back relevant elements when backtracking
element=numbers.pop(N)
numbers.insert(N+i,element)
numbers=[1,2,3]
permutations=[]
permute(numbers, permutations)
print "-----------"
for p in permutations:
print p
输出是这样的:
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
-----------
[1, 2, 3]
[1, 2, 3]
[1, 2, 3]
[1, 2, 3]
[1, 2, 3]
[1, 2, 3]
不完全符合我的预期...
答案 0 :(得分:3)
您追加到permutations
的所有项目实际上都是相同的列表。改变一个会改变所有其他人。最简单的解决方法是在添加列表之前复制列表,因此对numbers
的未来更改不会影响已经附加的结果。
def permute(numbers, permutations, N=0):
# base case
if N == len(numbers):
print numbers
permutations.append(numbers[:])
return
结果:
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]
---------
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]