Question

我正在尝试以递归的方式检索元素，这些元素求和为特定值（在这种情况下为9）。

在下图中的简化示例中，我有3个分支（值分别为3、4和5），并且最大可能总和为9，我希望得到以下输出：

[3,3,3] (= 9)
[3,4]   (= 7)
[3,5]   (8)
[4,3]   (7)
[4,4]   (8)
[4,5]   (9)
[5,3]   (8)

但是，我在下面的代码中的输出却大不相同：

i =  [3, 3, 3]
i =  [3, 3, 3]
i =  [3, 3, 3]
i =  [3, 3]
i =  [3, 3]
i =  [3, 3]
i =  [3]
i =  [3, 4]
i =  [3, 4]
i =  [3, 4]
i =  [3]
i =  [3, 5]
i =  [3, 5]
i =  [3, 5]
i =  [3]

我知道我可以使用itertools获得类似的输出，但是我的实际代码比这更复杂，因此我无法在其中使用iterool。这是我当前的代码：

import copy
maximum_sum = 9
branches = [3,5]
ar = []
big_ar = []

def recurse(summary, index):
    if summary + branches[index] <= maximum_sum:   
        summary += branches[index]
        ar.append(branches[index])

        for i in range(len(branches)):  
            recurse(summary, i)
            my_ar = copy.deepcopy(ar)
            big_ar.append(my_ar)
        del ar[-1]
    return (big_ar)

all_clusters = recurse(0, 0)
for i in all_clusters:
    print ('i = ', i)

Answer 1

maxsum = 9
branches = [3, 4, 5]

def recurse(maxsum):
    if maxsum < min_branch:
        yield []
    for b in branches:
        if maxsum >= b:
             for v in recurse(maxsum - b):
                   yield [*v, b]
        else:
            return

for i in recurse(9):
    print ('i = ', i)

Answer 2

解决方案-这就是我想要实现的目标。快速又肮脏，但是可以正常工作：

import copy
maximum_sum = 9
branches = [3,4,5]
ar = []
sum_ar = []

def recurse(summary, index):
    if summary + branches[index] <= maximum_sum:   
        summary += branches[index]
        ar.append(branches[index])

        for i in range(len(branches)): 
            recurse(summary, i)

        if summary + branches[index] > maximum_sum:    
            my_ar = copy.deepcopy(ar)
            sum_ar.append(my_ar)
        del ar[-1]
    return (sum_ar)

for i in range(len(branches)):
    all_clusters = recurse(0, i)

for i in all_clusters:
print ('i = ', i)

Python递归-总和堆栈

2 个答案: