生成所有连续子数组的算法

Question

使用以下输入，

[1, 2, 3, 4]

我正在尝试获得以下输出

[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [2], [2, 3], [2, 3, 4], [3], [3, 4], [4]]

目前我也做过这样的算法，但是时间复杂度不好。

def find(height):
    num1 = 0
    out = []
    for i in range(len(height)):
        num2 = 1
        for j in range(len(height)):
            temp = []
            for x in range(num1, num2):
                temp.append(height[x])
            num2 += 1
            if temp: out.append(temp)
        num1 += 1
    return out

有什么方法可以加快该算法的速度吗？

Answer 1

怎么样：

a = [1, 2, 3, 4]
l = len(a)
ret = []
for i in range(l):
    ll = i + 1
    while ll <= l:
        ret.append(a[i:ll])
        ll +=1
print(ret)

打印：

[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [2], [2, 3], [2, 3, 4], [3], [3, 4], [4]]

Answer 2

连续sub-sequences

他们在评论中指定的 OP 连续 sub-sequences.

select 一个连续的 sub-sequence 只需要 select 一个起始索引 i 和一个结束索引 j。然后我们可以简单地 return 切片 l[i:j].

def contiguous_subsequences(l):
  return [l[i:j] for i in range(0, len(l)) for j in range(i+1, len(l)+1)]

print(contiguous_subsequences([1,2,3,4]))
# [[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [2], [2, 3], [2, 3, 4], [3], [3, 4], [4]]

这个函数已经在包 more_itertools 中实现，它被调用 substrings:

import more_itertools
print(list(more_itertools.substrings([0, 1, 2])))
# [(0,), (1,), (2,), (0, 1), (1, 2), (0, 1, 2)]

Non-contiguous sub-sequences

为了完整性。

寻找可迭代对象的“幂集”是 itertool recipe:

import itertools
def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return itertools.chain.from_iterable(itertools.combinations(s, r) for r in range(len(s)+1))

它也在包中 more_itertools:

import more_itertools
print(list(more_itertools.powerset([1,2,3,4])))
# [(), (1,), (2,), (3,), (4,), (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4), (1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4), (1, 2, 3, 4)]

Answer 3

您可以简单地使用 列表理解 在一行 (O(N^2)) 中执行此操作，这比您现有的方法更快：

>>> x = [1,2,3,4]
>>> [x[i:j] for i in range(len(x)) for j in range(i+1,len(x)+1)] 
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [2], [2, 3], [2, 3, 4], [3], [3, 4], [4]]

运行时比较:

# your solution
>>> %timeit find(x)
9.23 µs ± 445 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

#itertools method suggested by 'stef' 
>>> %timeit list(more_itertools.substrings([1, 2, 3,4]))
3.18 µs ± 20.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

#List comprehension method
>>> %timeit [x[i:j] for i in range(len(x)) for j in range(i+1,len(x)+1)]
3.09 µs ± 27.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Answer 4

在这里，我使用 dynamic programming top - down 方法解决了这个问题。

这里的时间复杂度是O(N^2).

我不确定这个问题的时间复杂度是否可以进一步降低¯_(ツ)_/¯。

def find(arr):
    d = {}
    d[0] = []

    i = 1
    while (i <= len(arr)):
        d[i] = [] + d[i - 1]
        val = arr[i - 1]
        j = i - 1
        l = len(d[i - 1])
        while (j > 0):
            d[i].append(d[i - 1][l - j] + [val])
            j = j - 1

        d[i].append([val])
        i = i + 1

    return d[len(arr)]

input = [1, 2, 3, 4]
print(find(input))

输出：

[[1], [1, 2], [2], [1, 2, 3], [2, 3], [3], [1, 2, 3, 4], [2, 3, 4], [3, 4], [4]]

生成所有连续子数组的算法

Algorithm that generates all contiguous subarrays

python

algorithm

time-complexity

连续sub-sequences

Non-contiguous sub-sequences