我可以在不创建新的 list/array 的情况下展平数组吗？

Question

假设我有 arr = [[1], [0, 0], [2], [3], [0, 0], [4]]

我可以将其扁平化为 [1, 0, 0, 2, 3, 0, 0, 4] 而不必使用 itertools、map/reduce 或列表组合吗？我正在努力想办法做到这一点。谢谢。

这是我目前试过的，是一道leetcode题：https://leetcode.com/problems/duplicate-zeros/

class Solution:
    def duplicateZeros(self, arr: List[int]) -> None:
        """
        Do not return anything, modify arr in-place instead.
        
        """
        for ix in range(len(arr) - 2):
            if arr[ix] == 0:
                arr[ix] = [0, 0]
                del arr[-1]
            else:
                l = []
                l.append(arr[ix])
                arr[ix] = l

# Output when you return arr is [[1],[0, 0],[2],[3],[0, 0],[4]]

Answer 1

尝试一些递归：

def flatten(lst):
    ans = []
    for el in lst:
        if type(el) == list:
             for e in flatten(el):
                 ans.append(e)
        else:
            ans.append(el)
    return ans

它将展平任何维度的列表。

Answer 2

你可以简单地做：

arr = sum(arr, [])

你在这里做的是添加 arr 的可迭代元素，以空数组 [] 作为初始值。

Answer 3

我们这里也可以使用list copy([:])，解决问题：

class Solution:
    def duplicateZeros(self, A):
        """
        Do not return anything, modify arr in-place instead.
        """
        A[:] = [x for num in A for x in ([num] if num else [0, 0])][:len(A)]

此外，最佳解决方案是 N 运行时的顺序和恒定内存。 LeetCode here:

class Solution:
    def duplicateZeros(self, arr: List[int]) -> None:
        """
        Do not return anything, modify arr in-place instead.
        """

        possible_dups = 0
        length_ = len(arr) - 1

        # Find the number of zeros to be duplicated
        for left in range(length_ + 1):

            # Stop when left points beyond the last element in the original list
            # which would be part of the modified list
            if left > length_ - possible_dups:
                break

            # Count the zeros
            if arr[left] == 0:
                # Edge case: This zero can't be duplicated. We have no more space,
                # as left is pointing to the last element which could be included  
                if left == length_ - possible_dups:
                    arr[length_] = 0 # For this zero we just copy it without duplication.
                    length_ -= 1
                    break
                possible_dups += 1

        # Start backwards from the last element which would be part of new list.
        last = length_ - possible_dups

        # Copy zero twice, and non zero once.
        for i in range(last, -1, -1):
            if arr[i] == 0:
                arr[i + possible_dups] = 0
                possible_dups -= 1
                arr[i + possible_dups] = 0
            else:
                arr[i + possible_dups] = arr[i]

我可以在不创建新的 list/array 的情况下展平数组吗？

Can I flatten an array without making a new list/array?

python

arrays

list

flatten