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813. 最大平均值和的分组

思路：长度为

N

的数组，第一个分割位置在i位置时的平均值假设为

avg

， 剩余

N - i

长度的数组分割为

k - 1

个相邻子数组的最大分数平均值为

cur

。遍历数组中每一个

i

所在的位置，同时更新最大值

avg + cur


递归

class Solution {

    public double largestSumOfAverages(int[] nums, int k) {
        int N = nums.length;
        return process(nums, N,0, k);
    }
    private double process(int[] nums, int N, int index, int k) {
        if (k == 0) {
            return 0.0;
        }
        if (index >= N) {
            return 0.0;
        }
        //  k = 1时直接返回剩余数组的均值
        if (k == 1) {
            int sum = 0;
            for (int i = index; i < N; i++) {
                sum += nums[i];
            }
            return (sum * 1.0) / (N - index);
        }
        int sum = 0;
        double res = 0.0;
        // 枚举每个 i 所在的位置
        for (int i = index; i < N; i++) {
            sum += nums[i];
            double avg = (sum * 1.0) / (i - index + 1);
            // 剩余数组的最大均值
            double cur = process(nums, N, i + 1, k - 1);
            res = Math.max(res, avg + cur);
        }
        return res;
    }
}

递归会超时，因为存在许多重复计算，所以可以加缓存，也就是记忆化搜索

记忆化搜索

class Solution {
    public double largestSumOfAverages2(int[] nums, int k) {
        int N = nums.length;
        double[][] dp = new double[N + 1][k + 1];
        for (int i = 0; i < N + 1; i++) {
            Arrays.fill(dp[i], -1.0);
        }
        return process2(nums, N,0, k, dp);
    }

    private double process2(int[] nums, int N, int index, int k, double[][] dp) {
        if (dp[index][k] != -1.0) {
            return dp[index][k];
        }
        if (k == 0) {
            dp[index][k] = 0.0;
            return dp[index][k];
        }
        if (index >= N) {
            dp[index][k] = 0.0;
            return dp[index][k];
        }
        if (k == 1) {
            int sum = 0;
            for (int i = index; i < N; i++) {
                sum += nums[i];
            }
            dp[index][k] = (sum * 1.0) / (N - index);
            return dp[index][k];
        }
        int sum = 0;
        double res = 0.0;
        for (int i = index; i < N; i++) {
            sum += nums[i];
            double avg = (sum * 1.0) / (i - index + 1);
            double cur = process2(nums, N, i + 1, k - 1, dp);
            res = Math.max(res, avg + cur);
        }
        dp[index][k] = res;
        return dp[index][k];
    }
}

动态规划

将递归改写成动态规划

class Solution {
    public double dpWays(int[] nums, int k) {
        int N = nums.length;
        double[][] dp = new double[N + 1][k + 1];

        for (int i = N - 1; i >= 0; i--) {
            for (int j = 1; j < k + 1; j++) {
                if (j == 1) {
                    int sum = 0;
                    for (int l = i; l < N; l++) {
                        sum += nums[l];
                    }
                    dp[i][j] = (sum * 1.0) / (N - i);
                } else {
                    int sum = 0;
                    double res = 0.0;
                    for (int l = i; l < N; l++) {
                        sum += nums[l];
                        double avg = (sum * 1.0) / (l - i + 1);
                        double cur = dp[l + 1][j - 1];
                        res = Math.max(res, avg + cur);
                    }
                    dp[i][j] = res;
                }
            }
        }
        return dp[0][k];
    }
}