数据流中的中位数 牛客网 剑指Offer
题目描述如何得到一个数据流中的中位数？如果从数据流中读出奇数个数值，那么中位数就是所有数值排序之后位于中间的数值。如果从数据流中读出偶数个数值，那么中位数就是所有数值排序之后中间两个数的平均值。我们使用Insert()方法读取数据流，使用GetMedian()方法获取当前读取数据的中位数 class Solution: def __init__(self): self.left = [] self.right = [] self.count = 0 def Insert(self, num): if self.count & 1 == 0: self.left.append(num) else: self.right.append(num) self.count +=1 def GetMedian(self,x): if self.count == 1: return self.left[0] self.MaxHeap(self.left) self.MinHeap(self.right) if self.left[0] > self.right[0]: self.left[0],self.right[0] = self.right[0],self.left[0] self.MaxHeap(self.left) self.MinHeap(self.right) if self.count & 1 == 0: return (self.left[0] + self.right[0])/2.0 else: return self.left[0] def MaxHeap(self, alist): length = len(alist) if alist == None or length <= 0: return if length == 1: return alist for i in range(length//2-1, -1, -1): k = i; temp = alist[k]; heap = False while not heap and 2*k < length-1: index = 2*k+1 if index < length - 1: if alist[index] < alist[index + 1]: index += 1 if temp >= alist[index]: heap = True else: alist[k] = alist[index] k = index alist[k] = temp def MinHeap(self, alist): length = len(alist) if alist == None or length <= 0: return if length == 1: return alist for i in range(length//2-1, -1, -1): k = i; temp = alist[k]; heap = False while not heap and 2 * k < length - 1: index = 2 * k+1 if index < length - 1: if alist[index] > alist[index + 1]: index += 1 if temp <= alist[index]: heap = True else: alist[k] = alist[index] k = index alist[k] = temp
转载于:https://www.cnblogs.com/vercont/p/10210378.html
