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.
Examples:
[2,3,4] , the median is 3
[2,3], the median is (2 + 3) / 2 = 2.5
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.
For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.
Window position Median
--------------- -----
[1 3 -1] -3 5 3 6 7 1
1 [3 -1 -3] 5 3 6 7 -1
1 3 [-1 -3 5] 3 6 7 -1
1 3 -1 [-3 5 3] 6 7 3
1 3 -1 -3 [5 3 6] 7 5
1 3 -1 -3 5 [3 6 7] 6
Therefore, return the median sliding window as [1,-1,-1,3,5,6].
Note:
You may assume k is always valid, ie: k is always smaller than input array's size for non-empty array.
一刷
题解：用类似于347的思路，一个maxHeap, 一个minHeap
class Solution {
public double[] medianSlidingWindow(int[] nums, int k) {
int n = nums.length;//8
int m = n-k+1;//8-3+1=6
double[] res = new double[m];
PriorityQueue maxHeap = new PriorityQueue(k, Collections.reverseOrder());
PriorityQueue minHeap = new PriorityQueue(k);
for(int i=0; i=num) maxHeap.add(num);
else minHeap.add(num);
//the size are equal or maxHeap is larger than 1 element
if(minHeap.size()>maxHeap.size()){
maxHeap.add(minHeap.poll());
}else if(maxHeap.size()>minHeap.size()+1){
minHeap.add(maxHeap.poll());
}
if(i-k+1>=0){
if((k&1) == 1){//odd number
res[i-k+1] = maxHeap.peek();
}
else res[i-k+1] = maxHeap.peek()/2.0 + minHeap.peek()/2.0;
int toBeRemove = nums[i-k+1];
if(toBeRemove<=maxHeap.peek()) maxHeap.remove(toBeRemove);
else minHeap.remove(toBeRemove);
if(minHeap.size()>maxHeap.size()){
maxHeap.add(minHeap.poll());
}else if(maxHeap.size()>minHeap.size()+1){
minHeap.add(maxHeap.poll());
}
}
}
return res;
}
}