maximum sum of a subarray with at-least k elements.

// Returns maximum sum of a subarray with at-least
    // k elements.
    static int maxSumWithK(int a[], int n, int k)
    {
        // maxSum[i] is going to store maximum sum
        // till index i such that a[i] is part of the
        // sum.
        int maxSum[] = new int [n];
        maxSum[0] = a[0]; 

        // We use Kadane‘s algorithm to fill maxSum[]
        // Below code is taken from method 3 of
        // https://www.geeksforgeeks.org/largest-sum-contiguous-subarray/
        int curr_max = a[0];
        for (int i = 1; i < n; i++)
        {
            curr_max = Math.max(a[i], curr_max+a[i]);
            maxSum[i] = curr_max;
        } 

        // Sum of first k elements
        int sum = 0;
        for (int i = 0; i < k; i++)
            sum += a[i]; 

        // Use the concept of sliding window
        int result = sum;
        for (int i = k; i < n; i++)
        {
            // Compute sum of k elements ending
            // with a[i].
            sum = sum + a[i] - a[i-k]; 

            // Update result if required
            result = Math.max(result, sum); 

            // Include maximum sum till [i-k] also
            // if it increases overall max.
            result = Math.max(result, sum + maxSum[i-k]);
        }
        return result;
    }

原文地址：https://www.cnblogs.com/apanda009/p/9678608.html