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?
就是左边一个max heap, 右边一个 min Heap。
这里,要学的是,如何用 Collections.reverseOrder();
class MedianFinder { public: /** initialize your data structure here. */ MedianFinder() { } void addNum(int num) { // blindly added into the left leftMaxHeap.push(num); if(leftMaxHeap.size() > 1+ rightMinHeap.size() ){ int tmp = leftMaxHeap.top(); leftMaxHeap.pop(); rightMinHeap.push(tmp); } if(not rightMinHeap.empty() && leftMaxHeap.top() > rightMinHeap.top() ){ int bigger = leftMaxHeap.top(); leftMaxHeap.pop(); int smaller = rightMinHeap.top(); rightMinHeap.pop(); rightMinHeap.push(bigger); leftMaxHeap.push(smaller); } } double findMedian() { if(leftMaxHeap.size() > rightMinHeap.size()){ return leftMaxHeap.top(); }else{ return (leftMaxHeap.top() + rightMinHeap.top() )/2.0; } } private: priority_queue<int, vector<int>> leftMaxHeap; priority_queue<int, vector<int>,greater<int>> rightMinHeap; }; /** * Your MedianFinder object will be instantiated and called as such: * MedianFinder* obj = new MedianFinder(); * obj->addNum(num); * double param_2 = obj->findMedian(); */
Mistakes:
Amzaon面试中遇到了。
结果我没有检查
if(not rightMinHeap.empty() && leftMaxHeap.top() > rightMinHeap.top() ){
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