Leetcode: Sliding Window Median

 1 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.
 2 
 3 Examples: 
 4 [2,3,4] , the median is 3
 5 
 6 [2,3], the median is (2 + 3) / 2 = 2.5
 7 
 8 Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Your job is to output the median array for each window in the original array.
 9 
10 For example,
11 Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
12 
13 Window position                Median
14 ---------------               -----
15 [1  3  -1] -3  5  3  6  7       1
16  1 [3  -1  -3] 5  3  6  7       -1
17  1  3 [-1  -3  5] 3  6  7       -1
18  1  3  -1 [-3  5  3] 6  7       3
19  1  3  -1  -3 [5  3  6] 7       5
20  1  3  -1  -3  5 [3  6  7]      6
21 Therefore, return the median sliding window as [1,-1,-1,3,5,6].

方法1：Time Complexity O(NK)

暂时只有两个Heap的做法，缺点：In this problem, it is necessary to be able remove elements that are not necessarily at the top of the heap. PriorityQueue has logarithmic time remove top, but a linear time remove arbitrary element.

For a Heap:

remove():  Time Complexity is O(logN)

remove(Object): Time Complexity is O(N)

更好的有multiset的方法，但是还没有看到好的java version的

最大堆的简单定义方法：Collections.reverseOrder(), Returns a comparator that imposes the reverse of the natural ordering on a collection of objects 

 1 public class Solution {
 2     PriorityQueue<Double> high = new PriorityQueue();
 3     PriorityQueue<Double> low = new PriorityQueue(Collections.reverseOrder());
 4     
 5     
 6     public double[] medianSlidingWindow(int[] nums, int k) {
 7         double[] res = new double[nums.length-k+1];
 8         int index = 0;
 9 
10         for (int i=0; i<nums.length; i++) {
11             if (i >= k) remove(nums[i-k]);
12             add((double)nums[i]);
13             if (i >= k-1) {
14                 res[index++] = findMedian();
15             }
16         }
17         return res;
18     }
19     
20     public void add(double num) {
21         low.offer(num);
22         high.offer(low.poll());
23         if (low.size() < high.size()) {
24             low.offer(high.poll());
25         }
26     }
27     
28     public double findMedian() {
29         if (low.size() == high.size()) {
30             return (low.peek() + high.peek()) / 2.0;
31         }
32         else return low.peek();
33     }
34     
35     public void remove(double num) {
36         if (num <= findMedian()) {
37             low.remove(num);
38         }
39         else {
40             high.remove(num);
41         }
42         if (low.size() < high.size()) {
43             low.offer(high.poll());
44         }
45         else if (low.size() > high.size()+1) {
46             high.offer(low.poll());
47         }
48     }
49 }