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 permutations(head, tail=''):
if len(head) == 0:
print(tail)
else:
for i in range(len(head)):
permutations(head[:i] + head[i+1:], tail + head[i])
称为:
permutations('abc')
其他回答
此解决方案实现了一个生成器,以避免在内存中保留所有排列:
def permutations (orig_list):
if not isinstance(orig_list, list):
orig_list = list(orig_list)
yield orig_list
if len(orig_list) == 1:
return
for n in sorted(orig_list):
new_list = orig_list[:]
pos = new_list.index(n)
del(new_list[pos])
new_list.insert(0, n)
for resto in permutations(new_list[1:]):
if new_list[:1] + resto <> orig_list:
yield new_list[:1] + resto
以下代码是给定列表的就地排列,作为生成器实现。由于它只返回对列表的引用,因此不应在生成器外部修改列表。该解决方案是非递归的,因此使用了低内存。还可以很好地处理输入列表中元素的多个副本。
def permute_in_place(a):
a.sort()
yield list(a)
if len(a) <= 1:
return
first = 0
last = len(a)
while 1:
i = last - 1
while 1:
i = i - 1
if a[i] < a[i+1]:
j = last - 1
while not (a[i] < a[j]):
j = j - 1
a[i], a[j] = a[j], a[i] # swap the values
r = a[i+1:last]
r.reverse()
a[i+1:last] = r
yield list(a)
break
if i == first:
a.reverse()
return
if __name__ == '__main__':
for n in range(5):
for a in permute_in_place(range(1, n+1)):
print a
print
for a in permute_in_place([0, 0, 1, 1, 1]):
print a
print
from __future__ import print_function
def perm(n):
p = []
for i in range(0,n+1):
p.append(i)
while True:
for i in range(1,n+1):
print(p[i], end=' ')
print("")
i = n - 1
found = 0
while (not found and i>0):
if p[i]<p[i+1]:
found = 1
else:
i = i - 1
k = n
while p[i]>p[k]:
k = k - 1
aux = p[i]
p[i] = p[k]
p[k] = aux
for j in range(1,(n-i)/2+1):
aux = p[i+j]
p[i+j] = p[n-j+1]
p[n-j+1] = aux
if not found:
break
perm(5)
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)
```
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']