[leetcode] 295. Find Median from Data Stream @ python

最新推荐文章于 2021-08-19 14:54:55 发布

原创最新推荐文章于 2021-08-19 14:54:55 发布 · 899 阅读

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本文介绍了一种在数据流中实时计算中位数的方法，通过使用Python的bisect模块，确保数据列表始终保持升序，从而高效地找到中位数。此方法适用于处理大量连续输入的整数数据。

原题

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.

For example,
[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.

Example:

addNum(1)
addNum(2)
findMedian() -> 1.5
addNum(3)
findMedian() -> 2

Follow up:

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?

解法

使用bisect模块, bisect.insort_left使得每次插入数字后, 列表能保持升序排列, 然后求中值, (self.data[l//2] + self.data[(l-1)//2])/2.0 使得不管列表的元素是偶数个还是奇数个,结果都是中值.

代码

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()