边工作边刷题:70天一遍leetcode: day 45
Flatten Nested List Iterator
要点:题的本质是建立next和isInteger/getList/getInteger的关系,考点和tree的遍历类似:用stack存dfs的trace,这样可以backtrack到之前层的结点。
- 注意NestedInteger类的表示,清楚了表示很多题就迎刃而解了:NestedInteger表示一个结点:既可以是List也可以是Integer。什么类型由interface的get method返回值决定。而最外层object是List
,这样表示不可能以Integer为起点 - 为什么需要stack?本质上是tree的结构:遍历的时候如果当前元素是list,那么之后要,向更深遍历,并且要能返回的上层继续下一个元素。
- 如何unwind?对于list,相当于子树展开,这里就把list中元素反向push到stack里。因为list可能为空,这样就没有下一个元素可以iterate,所以要多层展开(也就是说,stack top是不能为list的)直到一个integer元素或者没有其他可遍历元素为止。因此,在init和next得到当前元素后,要用prepare来展开stack top的可能list
# """
# 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.stk = []
for i in xrange(len(nestedList)-1, -1, -1):
self.stk.append(nestedList[i])
self._prepare()
def _prepare(self):
while self.stk and not self.stk[-1].isInteger():
topList = self.stk.pop().getList()
for i in xrange(len(topList)-1, -1, -1):
self.stk.append(topList[i])
def next(self):
"""
:rtype: int
"""
topInt = self.stk.pop().getInteger()
self._prepare()
return topInt
def hasNext(self):
"""
:rtype: bool
"""
return self.stk
# Your NestedIterator object will be instantiated and called as such:
# i, v = NestedIterator(nestedList), []
# while i.hasNext(): v.append(i.next())