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341_FlattenNestedListIterator.py
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87 lines (69 loc) · 2.43 KB
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# coding: utf8
"""
题目链接: https://leetcode.com/problems/flatten-nested-list-iterator/description.
题目描述:
Given a nested list of integers, implement an iterator to flatten it.
Each element is either an integer, or a list -- whose elements may also be integers or other lists.
Example 1:
Given the list [[1,1],2,[1,1]],
By calling next repeatedly until hasNext returns false, the order of elements returned by next should be:
[1,1,2,1,1].
Example 2:
Given the list [1,[4,[6]]],
By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6].
"""
# """
# 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 = []
for n in nestedList[::-1]:
self.stack.append(n)
def next(self):
"""
:rtype: int
"""
return self.stack.pop()
def hasNext(self):
"""
:rtype: bool
"""
while self.stack:
# 内嵌列表, 出栈, 继续平铺展开
# 直至栈顶为整数
while self.stack and not self.stack[-1].isInteger():
s = self.stack.pop().getList()
for i in s[::-1]:
self.stack.append(i)
if self.stack and self.stack[-1].isInteger():
return True
return False
# Your NestedIterator object will be instantiated and called as such:
# i, v = NestedIterator(nestedList), []
# while i.hasNext(): v.append(i.next())