Leetcode: 408 sliding window median

The median is the number in the middle of an ordered sequence . If the length of the sequence is even , There is no middle number ; At this point, the median is the average of the two most intermediate numbers .

for example ：

[2,3,4], The median is  3
[2,3], The median is (2 + 3) / 2 = 2.5
Give you an array nums, There is a length of k From the left to the right . In the window k Number , Every time the window moves to the right 1 position . Your task is to find out the median number of elements in the new window after each window move , And output an array of them .

Example ：

give  nums = [1,3,-1,-3,5,3,6,7], as well as  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 , Returns the median array of the sliding window  [1,-1,-1,3,5,6].

Tips ：

You can assume  k  Always effective , namely ：k Always less than or equal to the number of elements of the input non empty array .
The error with the true value is within 10 ^ -5 The answer within will be regarded as the correct answer .

The first idea is to traverse the array , Take out a window array every time you traverse , Then get the results according to the rules and put them in the result array .

Code:

class Solution {
    public double[] medianSlidingWindow(int[] nums, int k) {
        
        double[] res = new double[nums.length-k+1];
        for (int i = 0; i + k <= nums.length; i++) {
             Array of sliding windows 
            double[] arr = new double[k];
            for (int j = 0; j < k; j++) {
                arr[j] = nums[j+i];
            }
            // stay arr Get the median in 
            Arrays.sort(arr);
            int mid = k / 2;
            res[i] = k % 2 == 0 ? (arr[mid-1] + arr[mid])/2 : arr[mid];
        }
        return res;
    }
}