341. Flatten Nested List Iterator
题目描述和难度
- 题目描述:
给出一个嵌套的整型列表。设计一个迭代器,遍历这个整型列表中的所有整数。
列表中的项或者为一个整数,或者是另一个列表。
示例 1:
给定列表 [[1,1],2,[1,1]]
,
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:
给定列表 [1,[4,[6]]]
,
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]
.
- 题目难度:中等。
- 英文网址:341. Flatten Nested List Iterator 。
- 中文网址:341. 扁平化嵌套列表迭代器 。
思路分析
求解关键:
参考解答
参考解答1
Python3 写法:
# """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
# class NestedInteger(object):
# def isInteger(self):
# """
# @return True if this NestedInteger holds a single integer, rather than a nested list.
# :rtype bool
# """
#
# def getInteger(self):
# """
# @return the single integer that this NestedInteger holds, if it holds a single integer
# Return None if this NestedInteger holds a nested list
# :rtype int
# """
#
# def getList(self):
# """
# @return the nested list that this NestedInteger holds, if it holds a nested list
# Return None if this NestedInteger holds a single integer
# :rtype List[NestedInteger]
# """
class NestedIterator(object):
def __init__(self, nestedList):
"""
Initialize your data structure here.
:type nestedList: List[NestedInteger]
"""
self.stack = nestedList[::-1]
# 临时存放值的地方
self.value = None
def next(self):
"""
:rtype: int
"""
result = self.hasNext()
if result:
ret = self.value
self.value = None
return ret
def hasNext(self):
"""
:rtype: bool
"""
if self.value is not None:
# 就不执行了,因为没有被取出来
return True
# 如果 self.value 为 None 的话
# 就要从 stack 中弹出元素了
while self.stack:
top = self.stack.pop()
# 如果栈中有元素,看看这个元素是什么
if top.isInteger():
# 是数字,就放在临时变量上
self.value = top.getInteger()
return True
else:
self.stack.extend(top.getList()[::-1])
return False
# Your NestedIterator object will be instantiated and called as such:
# i, v = NestedIterator(nestedList), []
# while i.hasNext(): v.append(i.next())
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