341.扁平化嵌套列表迭代器

难度：中等
题目描述：

思路总结：
方法一：用递归扁平化整个list
方法二：用栈在调用hasNext()的时候找到下一个元素
题解一：

# """
# 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
#        """
from collections import deque
class NestedIterator:
    def __init__(self, nestedList: [NestedInteger]):
        self.queue = deque()
        def visit(ls):
            for obj in ls:
                if obj.isInteger():
                    self.queue.append(obj.getInteger())
                else:
                    visit(obj.getList())
        visit(nestedList)
    
    def next(self) -> int:
        return self.queue.popleft()
            
    def hasNext(self) -> bool:
         return len(self.queue)>0
        
# Your NestedIterator object will be instantiated and called as such:
# i, v = NestedIterator(nestedList), []
# while i.hasNext(): v.append(i.next())

题解一结果：

来源：CSDN

作者：LotusQ

链接：https://blog.csdn.net/qq_30057549/article/details/103983349