题目描述：

LeetCode 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].

Python代码：

# """
# 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):
"""
:type nestedList: List[NestedInteger]
"""
self.stack = []
self.list = nestedList

def next(self):
"""
:rtype: int
"""
return self.stack.pop()

def hasNext(self):
"""
:rtype: bool
"""
while self.list or self.stack:
if not self.stack:
self.stack.append(self.list.pop(0))
while self.stack and not self.stack[-1].isInteger():
top = self.stack.pop().getList()
for e in top[::-1]:
self.stack.append(e)
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())

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