Data Stream Median
24%
Accepted
Numbers keep coming, return the median of numbers at every time a new number added.
Have you met this question in a real interview?
Yes
Example
For numbers coming list:
[1, 2, 3, 4, 5], return[1, 1, 2, 2, 3].
For numbers coming list:
[4, 5, 1, 3, 2, 6, 0], return[4, 4, 4, 3, 3, 3, 3].
For numbers coming list:
[2, 20, 100], return [2, 2, 20].
Challenge
Total run time in O(nlogn).
Clarification
What's the definition of Median? - Median is the number that in the middle of a sorted array. If there are n numbers in a sorted array A, the median is
A[(n - 1) / 2]. For example, if A=[1,2,3], median is 2. If A=[1,19], median is 1.
public class Solution {
/**
* @param nums: A list of integers.
* @return: the median of numbers
*/
public int[] medianII(int[] nums) {
// write your code here
if(nums == null || nums.length == 0) return nums;
int[] medians = new int[nums.length];
int size = 10;
Queue<Integer> minHeap = new PriorityQueue<Integer>(size, new Comparator<Integer>(){
public int compare(Integer a, Integer b){
return b - a;
}
});
Queue<Integer> maxHeap = new PriorityQueue<Integer>(size, new Comparator<Integer>(){
public int compare(Integer a, Integer b){
return a - b;
}
});
for(int i = 0; i < nums.length; i++){
if(minHeap.size() > maxHeap.size()){
if(nums[i] > minHeap.peek()){
maxHeap.offer(nums[i]);
} else {
maxHeap.offer(minHeap.poll());
minHeap.offer(nums[i]);
}
} else {
if(minHeap.isEmpty() || nums[i] <= maxHeap.peek()){
minHeap.offer(nums[i]);
} else {
minHeap.offer(maxHeap.poll());
maxHeap.offer(nums[i]);
}
}
medians[i] = minHeap.peek();
}
return medians;
}
}
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