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)java
暫時只有兩個Heap的作法,缺點: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.this
For a Heap:spa
remove(): Time Complexity is O(logN)code
remove(Object): Time Complexity is O(N)blog
更好的有multiset的方法,可是尚未看到好的java version的element
最大堆的簡單定義方法:Collections.reverseOrder(), Returns a comparator that imposes the reverse of the natural ordering on a collection of objects rem
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 }