Leetcode 341: Flatten Nested List Iterator

Code

# """
# 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]
        """
        # initialize a pair. 1st: the next element to explore, 
        # 2nd: the next index of that element to explore, if that element is a list-like NestedInteger
        self.stack = [[nestedList, 0]]
        
    def next(self):
        """
        :rtype: int
        """
        # hasNext() always makes sure the latest element on self.stack
        # is a list containing a non-list NestedInteger
        if self.hasNext():
            l, idx = self.stack.pop()
            return l.getInteger()

    def hasNext(self):
        """
        # hasNext() always makes sure the latest element on self.stack
        # is a list containing a non-list NestedInteger
        
        :rtype: bool
        """
        while self.stack:
            l, idx = self.stack[-1]
            if type(l) is NestedInteger and l.isInteger():
                return True
            else:
                if type(l) is NestedInteger and idx < len(l.getList()): 
                    self.stack[-1][1] += 1
                    self.stack.append([l.getList()[idx], 0])
                # the first added element in self.__init__ is a pure list    
                elif type(l) is list and idx < len(l):
                    self.stack[-1][1] += 1
                    self.stack.append([l[idx], 0])
                else:
                    # pop the element if the idx, the next position to explore, has already
                    # exceeded the length of the element
                    self.stack.pop()
        return False
        

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

 

Idea

Use a stack to maintain the elements to iterate. The next element to be iterated will be on the top of the stack. The stack is initialized as [nestedList, 0] in self.__init__(), as we are going to explore the first element in nestedList in the future.

In self.next(), we always first call self.hasNext(). The goal of self.hasNext() is to put the next integer on the top of the stack and also a boolean indicating whether there indeed exists the next integer to iterate.

In self.hasNext(), we peek the top element in the stack (line 55). If it is already an integer, return True and we don’t need to arrange the stack further. If it is a list of integers, we need to start exploring that list and put the next integer on the stack. Before that, we need to increase the top element’s index by one (line 60 and 64). This means that the next time when we visit this list, we should start exploring from the updated index.

The advantage of this algorithm is that it does not need to load all integers of the whole list into the stack at one time. Its space complexity is O(K), where K is the depth of the deepest nested list.

Reference: https://discuss.leetcode.com/topic/41870/real-iterator-in-python-java-c