Given an array of n integer with duplicate number, and a moving window(size k), move the window at each iteration from the start of the array, find the maximum number inside the window at each moving.
Have you met this question in a real interview?
Yes
Example
For array [1, 2, 7, 7, 8], moving window size k = 3. return [7, 7, 8]
At first the window is at the start of the array like this
[|1, 2, 7| ,7, 8] , return the maximum 7;
then the window move one step forward.
[1, |2, 7 ,7|, 8], return the maximum 7;
then the window move one step forward again.
[1, 2, |7, 7, 8|], return the maximum 8;
又是一道sliding window的题目。求中位数，和Find Median from Data Stream非常相似，可以想到用minheap和maxheap来解决是非常好的思路。
但是和data flow不一样， sliding window除了增还要删除，所以用hashheap来做比较好。
具体解法就是，把每次窗口的移动分解为一次增加元素，一次删除元素。因为有两个堆，判断在哪个堆中进行删除比较关键。因为题意，我的做法是maxheap总是存储和minheap一样数目或者多1的数字，这样中位数一直是maxheap的堆顶，则每次可以把要删除的元素和中位数比较，大则在minheap里面，小或者等于则在maxheap里删除。实际这种解法在lintcode中超时，但是如果多hashheap多写一个接口函数，contain(now),利用hash表进行检查，这种做法顺利通过，代码如下：
class Solution:
"""
@param nums: A list of integers.
@return: The median of element inside the window at each moving.
"""
def medianSlidingWindow(self, nums, k):
if len(nums) < k or not nums:
return []
minheap = HashHeap(‘min‘)
maxheap = HashHeap(‘max‘)
for i in xrange(k):
if i % 2:
maxheap.add(nums[i])
minheap.add(maxheap.pop())
else:
minheap.add(nums[i])
maxheap.add(minheap.pop())
median = []
for i in xrange(k, len(nums)):
median.append(maxheap.top())
if minheap.contain(nums[i - k]):
minheap.delete(nums[i - k])
maxheap.add(nums[i])
minheap.add(maxheap.pop())
else:
maxheap.delete(nums[i - k])
minheap.add(nums[i])
maxheap.add(minheap.pop())
median.append(maxheap.top())
return median
class Node(object):
"""
the type of class stored in the hashmap, in case there are many same heights
in the heap, maintain the number
"""
def __init__(self, id, num):
self.id = id #id means its id in heap array
self.num = num #number of same value in this id
class HashHeap(object):
def __init__(self, mode):
self.heap = []
self.mode = mode
self.size = 0
self.hash = {}
def top(self):
return self.heap[0] if len(self.heap) > 0 else 0
def contain(self, now):
if self.hash.get(now):
return True
else:
return False
def isempty(self):
return len(self.heap) == 0
def _comparesmall(self, a, b): #compare function in different mode
if a <= b:
if self.mode == ‘min‘:
return True
else:
return False
else:
if self.mode == ‘min‘:
return False
else:
return True
def _swap(self, idA, idB): #swap two values in heap, we also need to change
valA = self.heap[idA]
valB = self.heap[idB]
numA = self.hash[valA].num
numB = self.hash[valB].num
self.hash[valB] = Node(idA, numB)
self.hash[valA] = Node(idB, numA)
self.heap[idA], self.heap[idB] = self.heap[idB], self.heap[idA]
def add(self, now): #the key, height in this place
self.size += 1
if self.hash.get(now):
hashnow = self.hash[now]
self.hash[now] = Node(hashnow.id, hashnow.num + 1)
else:
self.heap.append(now)
self.hash[now] = Node(len(self.heap) - 1,1)
self._siftup(len(self.heap) - 1)
def pop(self): #pop the top of heap
self.size -= 1
now = self.heap[0]
hashnow = self.hash[now]
num = hashnow.num
if num == 1:
self._swap(0, len(self.heap) - 1)
self.hash.pop(now)
self.heap.pop()
self._siftdown(0)
else:
self.hash[now] = Node(0, num - 1)
return now
def delete(self, now):
self.size -= 1
hashnow = self.hash[now]
id = hashnow.id
num = hashnow.num
if num == 1:
self._swap(id, len(self.heap)-1) #like the common delete operation
self.hash.pop(now)
self.heap.pop()
if len(self.heap) > id:
self._siftup(id)
self._siftdown(id)
else:
self.hash[now] = Node(id, num - 1)
def parent(self, id):
if id == 0:
return -1
return (id - 1) / 2
def _siftup(self,id):
while abs(id -1)/2 < id : #iterative version
parentId = (id - 1)/2
if self._comparesmall(self.heap[parentId],self.heap[id]):
break
else:
self._swap(id, parentId)
id = parentId
def _siftdown(self, id): #iterative version
while 2*id + 1 < len(self.heap):
l = 2*id + 1
r = l + 1
small = id
if self._comparesmall(self.heap[l], self.heap[id]):
small = l
if r < len(self.heap) and self._comparesmall(self.heap[r], self.heap[small]):
small = r
if small != id:
self._swap(id, small)
else:
break
id = small
上述做法时间复杂度为O(nlogk)。但是这题需要求中位数，没有特别好的方法像Sliding Window Maximum这样找到O(n)的解法。就酱