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下面的代码返回提供的数字的所有可能的排列。递归中的python列表的行为
class Solution:
def permute(self, numbers, start, result):
if start == len(numbers):
print(numbers)
result.append(numbers[:])
return
for i in range(start, len(numbers)):
numbers[start], numbers[i] = numbers[i], numbers[start]
self.permute(numbers, start + 1, result)
numbers[start], numbers[i] = numbers[i], numbers[start]
def solution(self, numbers):
result = []
if not numbers or len(numbers) == 0:
return numbers
self.permute(numbers, 0, result)
return result
res1 = Solution().solution([1, 2, 3])
print(res1)
此实例的最终输出将是
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 2, 1], [3, 1, 2]]
但是当我稍微修改该置换功能,输出的是完全不同的
def permute(self, numbers, start, result):
if start == len(numbers):
print(numbers)
result.append(numbers) #changing this line
return
for i in range(start, len(numbers)):
numbers[start], numbers[i] = numbers[i], numbers[start]
self.permute(numbers, start + 1, result)
numbers[start], numbers[i] = numbers[i], numbers[start]
这给输出
[[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
该项目工程,当我使用
result.append([x for x in numbers])
或
result.append(numbers[:])
但不是当我使用
result.append(numbers)
有人可以帮助我理解为什么这是怎么回事?
作为卡罗利·霍瓦斯的答案,当你做数字[:]它为您创建一个新的数组。 – gipsy
供参考:'itertools.permutations'为你工作:) – Orelus