## Description

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:

Input: [[1,1],2,[1,1]]
Output: [1,1,2,1,1]
Explanation: By calling next repeatedly until hasNext returns false,
the order of elements returned by next should be: [1,1,2,1,1].


Example 2:

Input: [1,[4,[6]]]
Output: [1,4,6]
Explanation: By calling next repeatedly until hasNext returns false,
the order of elements returned by next should be: [1,4,6].


## Solutions

### 1. Stack

# Time: O(n)
# Space: 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.stack = nestedList[::-1]

def next(self) -> int:
return self.stack.pop().getInteger()

def hasNext(self) -> bool:
while self.stack:
top = self.stack[-1]
if top.isInteger():
return True
top = self.stack.pop()
self.stack += 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())

# 44/44 cases passed (64 ms)
# Your runtime beats 83.97 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (16.3 MB)