Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.
[2,3,4], the median is 3
[2,3], the median is (2 + 3)/2 = 2.5
Design a data structure that supports the following two operations:
void addNum(int num) – Add a integer number from the data stream to the data structure.
double findMedian() – Return the median of all elements so far.
findMedian() -> 1.5
findMedian() -> 2
If all integer numbers from the stream are between 0 and 100, how would you optimize it?
If 99% of all integer numbers from the stream are between 0 and 100, how would you optimize it?
Using the bisect module, bisect.insort_left keeps the list in ascending order each time a number is inserted, and then evaluates to the median, (self. data[l//2] + self.data[(l-1)//2])/2.0 so that the result is the median regardless of whether the list has even or odd elements. .
class MedianFinder: def __init__(self): """ initialize your data structure here. """ self.data =  def addNum(self, num): """ :type num: int :rtype: void """ bisect.insort_left(self.data, num) def findMedian(self): """ :rtype: float """ l = len(self.data) median = (self.data[l//2] + self.data[(l-1)//2])/2.0 return median # Your MedianFinder object will be instantiated and called as such: # obj = MedianFinder() # obj.addNum(num) # param_2 = obj.findMedian()
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