题目:有一个源源不断地吐出整数的数据流,假设你有足够的空间来保存吐出的数。请设计一个名叫MedianHolder的结构,MedianHolder可以随时取得之前吐出所有数的中位数。
解析:关于此问题的主要解题思路为建立大根堆和小根堆,大根堆用来存储较小的数,小根堆用来存储较大的数,在读入数据的过程中要进行大根堆和小根堆的调整,使两者所保存的数据量的差值不大于2,主要的步骤如下:
建堆完成后,如果大根堆和小根堆的size相等,那么取两个堆得堆顶元素的平均数即可,如果不相等,那么取size较大的堆的堆顶元素即可。
首先,我们利用Python自带的heapq模块分别建立大根堆和小根堆。
import heapqclass BigHeap():def __init__(self):self.arr = list()def heap_insert(self, val):heapq.heappush(self.arr, -val)def heapify(self):heapq.heapify(self.arr)def heap_pop(self):return -heapq.heappop(self.arr)def get_top(self):if not self.arr:returnreturn -self.arr[0]
import heapqclass SmallHeap():def __init__(self):self.arr = list()def heap_insert(self, val):heapq.heappush(self.arr, val)def heapify(self):heapq.heapify(self.arr)def heap_pop(self):return heapq.heappop(self.arr)def get_top(self):if not self.arr:returnreturn self.arr[0]
接下来我们设计MedianHolder结构来获取数据流中的中位数。
class MedianHolder():def __init__(self):self.bigHeap = BigHeap()self.smallHeap = SmallHeap()def addNum(self, num):if len(self.bigHeap.arr) == 0:self.bigHeap.heap_insert(num)returnif _top() >= num:self.bigHeap.heap_insert(num)else:if len(self.smallHeap.arr) == 0:self.smallHeap.heap_insert(num)returnif _top() > num:self.bigHeap.heap_insert(num)else:self.smallHeap.heap_insert(difyTwoHeapSize()def getMedian(self):smallHeapSize = len(self.smallHeap.arr)bigHeapSize = len(self.bigHeap.arr)if smallHeapSize + bigHeapSize == 0:return NonesmallHeapHead = _top()bigHeapHead = _top()if (smallHeapSize + bigHeapSize) %2 == 0:return (smallHeapHead+bigHeapHead)/2else:return smallHeapHead if smallHeapSize > bigHeapSize else bigHeapHeaddef modifyTwoHeapSize(self):smallHeapSize = len(self.smallHeap.arr)bigHeapSize = len(self.bigHeap.arr)if smallHeapSize == bigHeapSize + 2:self.bigHeap.heap_insert(self.smallHeap.heap_pop())if bigHeapSize == smallHeapSize + 2:self.smallHeap.heap_insert(self.bigHeap.heap_pop())
测试用例如下:
if __name__ == '__main__':arr = [68,51,42,92,13,46,24,58,62,72,32]medianHold = MedianHolder()for i in range(len(arr)):medianHold.addNum(arr[i])Median())
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