341. 扁平化嵌套列表迭代器 - medium
给你一个嵌套的整型列表。请你设计一个迭代器,使其能够遍历这个整型列表中的所有整数。
列表中的每一项或者为一个整数,或者是另一个列表。其中列表的元素也可能是整数或是其他列表。
示例 1:
输入: [[1,1],2,[1,1]]
输出: [1,1,2,1,1]
解释: 通过重复调用 next 直到 hasNext 返回 false,next 返回的元素的顺序应该是: [1,1,2,1,1]
。
示例 2:
输入: [1,[4,[6]]]
输出: [1,4,6]
解释: 通过重复调用 next 直到 hasNext 返回 false,next 返回的元素的顺序应该是: [1,4,6]
。
递归做 DFS。利用 Python 原生的迭代器,避免额外的存储空间,但是其实也需要保存栈的信息。时间复杂度 O(n)
# """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
#class NestedInteger:
# def isInteger(self) -> bool:
# """
# @return True if this NestedInteger holds a single integer, rather than a nested list.
# """
#
# def getInteger(self) -> int:
# """
# @return the single integer that this NestedInteger holds, if it holds a single integer
# Return None if this NestedInteger holds a nested list
# """
#
# def getList(self) -> [NestedInteger]:
# """
# @return the nested list that this NestedInteger holds, if it holds a nested list
# Return None if this NestedInteger holds a single integer
# """
class NestedIterator:
def __init__(self, nestedList: [NestedInteger]):
self.g = iterator(nestedList)
self.n = None
def next(self) -> int:
v = self.n
if v is None:
return next(self.g)
self.n = None
return v
def hasNext(self) -> bool:
if self.n is not None:
return True
try:
self.n = next(self.g)
except StopIteration:
return False
return True
def iterator(nl):
for n in nl:
if n.isInteger():
yield n.getInteger()
else:
for i in iterator(n.getList()):
yield i
# Your NestedIterator object will be instantiated and called as such:
# i, v = NestedIterator(nestedList), []
# while i.hasNext(): v.append(i.next())
贴合题目中给出的调用方式,可以更精简一些:
class NestedIterator:
def __init__(self, nestedList: [NestedInteger]):
self.g = iterator(nestedList)
self.n = None
def next(self) -> int:
v, self.n = self.n, None
return v
def hasNext(self) -> bool:
try:
self.n = next(self.g)
except StopIteration:
return False
return True
def iterator(nl):
for n in nl:
if n.isInteger():
yield n.getInteger()
else:
for i in iterator(n.getList()):
yield i