---
description: Heap | Priority Queue
---

# Leetcode 295. Find Median from Data Stream



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:**

1. If all integer numbers from the stream are between 0 and 100, how would you optimize it?
2. If 99% of all integer numbers from the stream are between 0 and 100, how would you optimize it?

## **Solution:**&#x20;

[English Version in Youtube](https://youtu.be/sSUCgilaVJ4)

[中文版解答Youtube Link](https://youtu.be/ByZgcvujtSA)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV16v411s7KH/)

* Use 2 priority queues. One is maximum heap, one is minimum heap.
* Put smaller elements in maximum heap, put greater elements in minimum heap.
* Try to keep the size same. If total size is odd then we can put extra element in minimum heap.

```
class MedianFinder {
    priority_queue<int, vector<int>, std::less<int>> maximum_heap;
    priority_queue<int, vector<int>, std::greater<int>> minimum_heap;
public:
    /** initialize your data structure here. */
    MedianFinder() {
        
    }
    
    void addNum(int num) {
        minimum_heap.push(num);
        
        maximum_heap.push(minimum_heap.top());
        minimum_heap.pop();
        
        if (minimum_heap.size() < maximum_heap.size()) {
            minimum_heap.push(maximum_heap.top());
            maximum_heap.pop();
        }
    }
    
    double findMedian() {
        if (maximum_heap.size() == minimum_heap.size()) {
            return (minimum_heap.top()+maximum_heap.top()) / 2.0;
        }
        return minimum_heap.top();
    }
};
```
