643. Maximum Average Subarray

Given an array consisting of \(n\) integers, find the contiguous subarray of given length \(k\) that has the maximum average value. And you need to output the maximum average value.

Example 1:

Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75

Note:

1. 1 <= k <= n <= 30,000.
2. Elements of the given array will be in the range [-10,000, 10,000].

自己方法1 Brute Force:
窗口k = 4, 从数组左到右滑动, 代码省略.
\(O(n*k)\) time, \(O(1)\) extra space.

自个方法2 改进版的 Brute Force:
想法如下:

k = 4 的情形

1	12	-5	-6	50	3
^			 ^

1	12	-5	-6	50	3
	^			^
明显的, 第一行的 sum  + 50  -  1 就是第二行的 sum
					A[i]  A[i-k]

\(O(n)\) time, \(O(1)\) extra space.

double findMaxAverage(vector<int>& A, int k) {
    int i, sum = 0, max = INT_MIN;
    for (i = 0; i < k; i++) sum += A[i];
    max = max > sum ? max : sum;
    for (i = k; i < A.size(); i++) {
        sum = sum + A[i] - A[i - k];
        max = max > sum ? max : sum;
    }
    return 1.0 * max / k;
}