The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value, and the median is the mean of the two middle values.
arr = [2,3,4], the median is 3.arr = [2,3], the median is (2 + 3) / 2 = 2.5.Implement the MedianFinder class:
MedianFinder() initializes the MedianFinder object.void addNum(int num) adds the integer num from the data stream to the data structure.double findMedian() returns the median of all elements so far. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input ["MedianFinder", "addNum", "addNum", "findMedian", "addNum", "findMedian"] [[], [1], [2], [], [3], []] Output [null, null, null, 1.5, null, 2.0] Explanation MedianFinder medianFinder = new MedianFinder(); medianFinder.addNum(1); // arr = [1] medianFinder.addNum(2); // arr = [1, 2] medianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2) medianFinder.addNum(3); // arr[1, 2, 3] medianFinder.findMedian(); // return 2.0
Constraints:
-105 <= num <= 105findMedian.5 * 104 calls will be made to addNum and findMedian.
Follow up:
[0, 100], how would you optimize your solution?99% of all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?class MedianFinder {
priority_queue<long> small, large;
public:
void addNum(int num) {
small.push(num);
large.push(-small.top());
small.pop();
if (small.size() < large.size()) {
small.push(-large.top());
large.pop();
}
}
double findMedian() {
return small.size() > large.size()
? small.top()
: (small.top() - large.top()) / 2.0;
}
};
/**
* Your MedianFinder object will be instantiated and called as such:
* MedianFinder* obj = new MedianFinder();
* obj->addNum(num);
* double param_2 = obj->findMedian();
*/