版权声明:版权归个人所有,未经博主允许,禁止转载 https://blog.csdn.net/danspace1/article/details/86619996
原题
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].
解法
双向队列. 我们在初始化的时候就将nestedList转换成deque, 利用题目中定义的方法, 我们遍历nestedList, 如果元素是整数, 那么直接加到deque里, 如果元素是nestedList, 那么递归求解.
next函数就是每次在左边删除一个数字, hasNext函数就是查看deque的长度是否大于0.
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):
"""
Initialize your data structure here.
:type nestedList: List[NestedInteger]
"""
def flatten(nestedList):
for l in nestedList:
if l.isInteger():
self.q.append(l)
else:
flatten(l.getList())
self.q = collections.deque()
flatten(nestedList)
def next(self):
"""
:rtype: int
"""
return self.q.popleft()
def hasNext(self):
"""
:rtype: bool
"""
return len(self.q)> 0
# Your NestedIterator object will be instantiated and called as such:
# i, v = NestedIterator(nestedList), []
# while i.hasNext(): v.append(i.next())