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
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。
其他回答
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。
对于Python 2.6及以上版本:
import itertools
itertools.permutations([1, 2, 3])
这将作为生成器返回。使用列表(排列(xs))作为列表返回。
from typing import List
import time, random
def measure_time(func):
def wrapper_time(*args, **kwargs):
start_time = time.perf_counter()
res = func(*args, **kwargs)
end_time = time.perf_counter()
return res, end_time - start_time
return wrapper_time
class Solution:
def permute(self, nums: List[int], method: int = 1) -> List[List[int]]:
perms = []
perm = []
if method == 1:
_, time_perm = self._permute_recur(nums, 0, len(nums) - 1, perms)
elif method == 2:
_, time_perm = self._permute_recur_agian(nums, perm, perms)
print(perm)
return perms, time_perm
@measure_time
def _permute_recur(self, nums: List[int], l: int, r: int, perms: List[List[int]]):
# base case
if l == r:
perms.append(nums.copy())
for i in range(l, r + 1):
nums[l], nums[i] = nums[i], nums[l]
self._permute_recur(nums, l + 1, r , perms)
nums[l], nums[i] = nums[i], nums[l]
@measure_time
def _permute_recur_agian(self, nums: List[int], perm: List[int], perms_list: List[List[int]]):
"""
The idea is similar to nestedForLoops visualized as a recursion tree.
"""
if nums:
for i in range(len(nums)):
# perm.append(nums[i]) mistake, perm will be filled with all nums's elements.
# Method1 perm_copy = copy.deepcopy(perm)
# Method2 add in the parameter list using + (not in place)
# caveat: list.append is in-place , which is useful for operating on global element perms_list
# Note that:
# perms_list pass by reference. shallow copy
# perm + [nums[i]] pass by value instead of reference.
self._permute_recur_agian(nums[:i] + nums[i+1:], perm + [nums[i]], perms_list)
else:
# Arrive at the last loop, i.e. leaf of the recursion tree.
perms_list.append(perm)
if __name__ == "__main__":
array = [random.randint(-10, 10) for _ in range(3)]
sol = Solution()
# perms, time_perm = sol.permute(array, 1)
perms2, time_perm2 = sol.permute(array, 2)
print(perms2)
# print(perms, perms2)
# print(time_perm, time_perm2)
```
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)