295. Find Median from Data Stream

Description

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?

Problem URL

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Solution

设计一个数据结构，能够存贮一系列的数，并且快速找到它们的中位数。

Using two priority queue(heap) to solve this problem. left is a max heap which stores left part, and right is a min heap which stores right part. When a new number comes in, store to right first then store right.poll() to left. If right.size() < left.size(), poll() left to right to make sure median is in right part when number of numbers is odd.

Code

class MedianFinder {
    Queue<Integer> left;
    Queue<Integer> right;
    /** initialize your data structure here. */
    public MedianFinder() {
        left = new PriorityQueue<>();
        right = new PriorityQueue<>(Collections.reverseOrder());
    }
    
    public void addNum(int num) {
        right.offer(num);
        left.offer(right.poll());
        if (right.size() < left.size()){
            right.offer(left.poll());
        }
    }
    
    public double findMedian() {
        if (right.size() == left.size()){
            return (right.peek() + left.peek()) / 2.0;
        }
        else{
            return right.peek();
        }
    }
}

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder obj = new MedianFinder();
 * obj.addNum(num);
 * double param_2 = obj.findMedian();
 */

Time Complexity: O()
Space Complexity: O()

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