Hard
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 <= 105
findMedian
.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?import HeapModule
class MedianFinder {
var leftHeap: Heap<Int>
var rightHeap: Heap<Int>
init() {
leftHeap = []
rightHeap = []
}
func addNum(_ num: Int) {
leftHeap.insert(num)
while let rightMin = rightHeap.min, rightMin < leftHeap.max! {
leftHeap.insert(rightHeap.removeMin())
rightHeap.insert(leftHeap.removeMax())
}
let mid = (leftHeap.count + rightHeap.count) / 2
while rightHeap.count < mid {
rightHeap.insert(leftHeap.removeMax())
}
}
func findMedian() -> Double {
let count = leftHeap.count + rightHeap.count
if count % 2 == 0 {
return Double(leftHeap.max! + rightHeap.min!) / 2
} else {
return Double(leftHeap.max!)
}
}
}
/*
* Your MedianFinder object will be instantiated and called as such:
* let obj = MedianFinder()
* obj.addNum(num)
* let ret_2: Double = obj.findMedian()
*/