def nested_depth(L):
"""
>>> nested_depth([1, 2, 3])
1
>>> nested_depth([1, [2, 3], 4])
2
"""
return (1 + max([nested_depth(x) for x in L])
if isinstance(L, list) else 0)
如何将此递归代码转换为迭代代码?即我们必须使用循环,我们不允许自己调用函数。
由于
答案 0 :(得分:3)
有一种将递归转换为迭代的“通用”方法。您必须引入并维护自己的堆栈,以维护在递归中执行的函数帧堆栈为您维护的状态。例如:
def max_nested_depth(items):
assert type(items) == list
max_depth = depth = 1
# Use a stack to maintain the position in each layer of lists.
stack = [(0, items)]
while stack: # Continue until the stack is empty.
# Iterate each layer until the index exceeds the end of the list.
while stack[-1][0] < len(stack[-1][1]):
pos, items = stack[-1]
stack[-1] = (pos + 1, items)
if type(items[pos]) == list:
# Here's the recursion step.
stack.append((0, items[pos]))
depth += 1
max_depth = max(max_depth, depth)
stack.pop()
depth -= 1
return max_depth
答案 1 :(得分:2)
>>> def maxDepth(L):
... answer = 1
... while L:
... if not any(isinstance(i, list) for i in L):
... return answer
... answer += 1
... L = list(itertools.chain.from_iterable([i for i in L if isinstance(i, list)]))
...
>>> L = [1,2,3]
>>> maxDepth(L)
1
>>> L = [1,[2,3],4]
>>> maxDepth(L)
2
答案 2 :(得分:1)
与@ inspectorG4dget
相同的想法def max_depth(L):
ans = 0
while L:
L = [j for i in L for j in (i if isinstance(i, list) else [])]
ans += 1
return ans