动态规划-Largest Sum of Averages

2018-07-12 23:21:53

问题描述：

问题求解：

dp[i][j] : 以ai结尾的分j个部分得到的最大值

dp[i][j] = max{dp[k][j - 1] + (a_k+1 + ... + a_i) / (i - k)} k = [j - 2, i - 1]

    public double largestSumOfAverages(int[] A, int K) {
        double[][] dp = new double[A.length][K + 1];
        int sum = 0;
        for (int i = 0; i < A.length; i++) {
            sum += A[i];
            dp[i][1] = sum * 1.0 / (i + 1);
        }
        for (int j = 2; j < K + 1; j++) {
            for (int i = j - 1; i < A.length; i++) {
                for (int k = j - 2; k < i; k++) {
                    sum = 0;
                    int temp = k + 1;
                    while (temp <= i) sum += A[temp++];
                    double avg = sum * 1.0 / (i - k);
                    dp[i][j] = Math.max(dp[i][j], dp[k][j - 1] + avg);
                }
            }
        }
        return dp[A.length - 1][K];
    }