341 Flatten Nested List Iterator

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].

The Idea: A nestedList is nothing but a tree. Naturally, a tree is recursively traversed, but we don't have that luxiery here because call stack has to be saved, in a sense. Any nestedList can be represented as a cons cell (lisp idea). For example, the tree below represents the following list: (1 (2 6 7 8) 3 (4 (9 12)) (5 10 11)). As we can see, following a preorder traversal through this tree will reveal the correct order of the elements within the nested list. A preorder traversal can be accomplished iteratively using a stack.

The approach to this problem is the same, except that we'll need to store an iterator for every sublist.

Any element within a nestedList can be either of two things: an integer or a list. If the element is an integer, we're ok to output it. Otherwise, if it is a list, we need to push this list onto the stack, and initialize it's own iterator (starting from 0).

Complexity: O(N) time where N is the total number of element in the list (including the nested lists), and O(|deepest nestest list|) space

# """
# 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.s = [[nestedList, 0]]

    def next(self):
        """
        :rtype: int
        """
        l, i = self.s[-1]
        self.s[-1][1] += 1
        return l[i].getInteger()

    def hasNext(self):
        """
        :rtype: bool
        """
        # get to the point where we can find an integer
        while self.s:
            nl, i = self.s[-1]

            # the current list is exhausted
            if i == len(nl):
                self.s.pop()

            # current admidst an iteration through nonexhausted list
            elif nl[i].isInteger():
                return True

            # otherwise next element must be a list
            # add this to the stack so we can backtrack to it later
            else:
                self.s[-1][1] += 1
                self.s.append([nl[i].getList(), 0])

        # reached back to the root, iteration is complete
        return False




        # Your NestedIterator object will be instantiated and called as such:
        # i, v = NestedIterator(nestedList), []
        # while i.hasNext(): v.append(i.next())