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