问题描述
假设我有arr = [[1],[0,0],[2],[3],[4]]
是否可以不必使用itertools，map / reduce或list comp将其展平为[1,2,3,4]？我正在努力想办法做到这一点。谢谢。
这是我到目前为止尝试过的，这是一个leetcode问题：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],[4]]
解决方法
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
它将拼合任意维度的列表。
,您可以轻松地做到：
arr = sum(arr,[])
您在这里要做的是添加arr的可迭代元素，并将空数组[]作为初始值。
- 我们也可以在此处使用list copy（
[:]）来解决问题：
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，最佳解决方案是具有恒定内存的N个运行时的顺序。 LeetCode here已经解决了这个问题：
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]