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.
 For example, for
arr = [2,3,4]
, the median is3
.  For example, for
arr = [2,3]
, the median is(2 + 3) / 2 = 2.5
.
Implement the MedianFinder class:
MedianFinder()
initializes theMedianFinder
object.void addNum(int num)
adds the integernum
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:
10^{5} <=
num
<= 10^{5} There will be at least one element in the data structure before calling
findMedian
.  At most
5 * 10^{4}
calls will be made toaddNum
andfindMedian
.
Follow up:
 If all integer numbers from the stream are in the range
[0, 100]
, how would you optimize your solution?  If
99%
of all integer numbers from the stream are in the range[0, 100]
, how would you optimize your solution?
insort
would beO(n)
because all the elements in the array may need to be shifted after insertion

