如何在Python中返回递归函数的列表

Question

我试图从函数返回一个字符串列表，该函数计算所有可能的排列，而不连续0

为此，我运行了一个可以工作的递归函数，但我需要创建一个包含结果的列表

# Function to print all n–digit binary strings without any consecutive 0's
def countStrings(n, out="", last_digit=0):
    # if the number becomes n–digit, print it
    
    if n == 0:
        print(out)
        return
    
    # append 0 to the result and recur with one less digit
    countStrings(n - 1, out + '1', 0)
    
 
    # append 1 to the result and recur with one less digit
    # only if the last digit is 0
    if last_digit == 0:
        countStrings(n - 1, out + '0', 1)

当我运行它时，例如：a = countStrings(3)，它打印所有可能的变量，但变量“a”返回为“None”：

结果:

111
110
101
011
010

type(a): Nonetype

我试图在某些地方插入一个附加，但没有结果

我不知道我错过了什么

Answer 1

如果不想使用return，可以创建global参数，并在函数完成后调用它以执行：

res = []

def countStrings(n, out="", last_digit=0):
    global res
    # if the number becomes n–digit, print it
    if n == 0:
        print(out)
        res.append(out)
        return 
    
    # append 0 to the result and recur with one less digit
    countStrings(n - 1, out + '1', 0)
    
 
    # append 1 to the result and recur with one less digit
    # only if the last digit is 0
    if last_digit == 0:
        countStrings(n - 1, out + '0', 1)

调用print(res)将为您提供：

['111', '110', '101', '011', '010']

Answer 2

这就是你要找的吗

# Function to print all n–digit binary strings without any consecutive 0's
def countStrings(n, out="", last_digit=0, acc=[]):
    # if the number becomes n–digit, print it
    
    if n == 0:
        print(out)
        acc.append(out)
        return acc
    
    # append 0 to the result and recur with one less digit
    countStrings(n - 1, out + '1', 0, acc)
    
 
    # append 1 to the result and recur with one less digit
    # only if the last digit is 0
    if last_digit == 0:
        countStrings(n - 1, out + '0', 1, acc)
    return acc

acc = countStrings(3)
print('acc', acc)

输出：

111
110
101
011
010
('acc', ['111', '110', '101', '011', '010'])

Answer 3

解决方案A

我建议将该计划的关注点分为多个功能：

def binaries(n):
  for m in range(2 ** n):
    yield f"{m:>0{n}b}"

print(list(binaries(3)))

['000', '001', '010', '011', '100', '101', '110', '111']

现在我们可以写pairwise，它在一个iterable上成对迭代-

from itertools import tee, islice

def pairwise(t):
  a, b = tee(t)
  yield from zip(a, islice(b, 1, None))

print(list(pairwise("011001")))

[('0', '1'), ('1', '1'), ('1', '0'), ('0', '0'), ('0', '1')]

现在很容易将solution实现为所有binaries大小n的组合，其中pairwise对任何特定二进制的分析不包含两个相邻的零-

def solution(n):
  for b in binaries(n):
    for p in pairwise(b):
      if p == ('0','0'):
        break
    else:
      yield b

print(list(solution(3)))

['010', '011', '101', '110', '111']

如果所有人都只需要一个解决方案的计数，我们可以将count_solutions写为solutions-

def count_solutions(n):
  return len(list(solution(n)))

print(count_solutions(3))

5

让我们来看一个更大的例子，n=8-

print(list(solutions(8)))
print(count_solutions(8))

['01010101', '01010110', '01010111', '01011010', '01011011', '01011101', '01011110', '01011111', '01101010', '01101011', '01101101', '01101110', '01101111', '01110101', '01110110', '01110111', '01111010', '01111011', '01111101', '01111110', '01111111', '10101010', '10101011', '10101101', '10101110', '10101111', '10110101', '10110110', '10110111', '10111010', '10111011', '10111101', '10111110', '10111111', '11010101', '11010110', '11010111', '11011010', '11011011', '11011101', '11011110', '11011111', '11101010', '11101011', '11101101', '11101110', '11101111', '11110101', '11110110', '11110111', '11111010', '11111011', '11111101', '11111110', '11111111']
55

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解决方案B

我一直在思考这个问题，我不喜欢上面的解决方案中的数字在比较之前如何转换成字符串

首先我们写bitwise-

def bitwise(n, e):
  if e == 0:
    return
  else:
    yield from bitwise(n >> 1, e - 1)
    yield n & 1

for n in range(2 ** 3):
  print(tuple(bitwise(n, 3)))

(0, 0, 0)
(0, 0, 1)
(0, 1, 0)
(0, 1, 1)
(1, 0, 0)
(1, 0, 1)
(1, 1, 0)
(1, 1, 1)

然后pairwise-

from itertools import tee, islice

def pairwise(t):
  a, b = tee(t)
  yield from zip(a, islice(b, 1, None))

for n in range(2 ** 3):
  print(bin(n), list(pairwise(bitwise(n, 3))))

0b0 [(0, 0), (0, 0)]
0b1 [(0, 0), (0, 1)]
0b10 [(0, 1), (1, 0)]
0b11 [(0, 1), (1, 1)]
0b100 [(1, 0), (0, 0)]
0b101 [(1, 0), (0, 1)]
0b110 [(1, 1), (1, 0)]
0b111 [(1, 1), (1, 1)]

最后solution-

def solution(e):
  for n in range(2 ** e):
    for p in pairwise(bitwise(n, e)):
      if p == (0, 0):
        break
    else:
      yield n

sln = list(map(bin, solution(3)))
print(sln)
print(len(sln))

['0b10', '0b11', '0b101', '0b110', '0b111']
5

和n=8-

sln = list(map(bin, solution(8)))
print(sln)
print(len(sln))

['0b1010101', '0b1010110', '0b1010111', '0b1011010', '0b1011011', '0b1011101', '0b1011110', '0b1011111', '0b1101010', '0b1101011', '0b1101101', '0b1101110', '0b1101111', '0b1110101', '0b1110110', '0b1110111', '0b1111010', '0b1111011', '0b1111101', '0b1111110', '0b1111111', '0b10101010', '0b10101011', '0b10101101', '0b10101110', '0b10101111', '0b10110101', '0b10110110', '0b10110111', '0b10111010', '0b10111011', '0b10111101', '0b10111110', '0b10111111', '0b11010101', '0b11010110', '0b11010111', '0b11011010', '0b11011011', '0b11011101', '0b11011110', '0b11011111', '0b11101010', '0b11101011', '0b11101101', '0b11101110', '0b11101111', '0b11110101', '0b11110110', '0b11110111', '0b11111010', '0b11111011', '0b11111101', '0b11111110', '0b11111111']
55

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建议阅读

如果您从未在Python中看到过for..else循环，请read the docs了解这个奇妙的构造