Today's headlines-dynamic programming-maximum interval

Title description:

Given an array sequence, it is required to select an interval so that the interval is the largest value among all the intervals calculated as follows:

The smallest number in the interval * the sum of all the numbers in the interval and the final program output the calculated maximum value. There is no need to output the collective interval.

For a given sequence [6 2 1], according to the above formula, all the calculated values ​​that can be selected for each interval can be obtained.

[6]=6*6=36;

[2]=2*2=4;

[1]=1*1=1;

[6,2]=2*8=16;

[2,1]=1*3=3;

[6,2,1]=1*9=9;

From the above calculation, it can be seen that the selected interval [6], the calculated value is 36, and the program outputs 36.

All numbers in the interval are in the range of [0,100].

Enter a description:

Enter the length of the array sequence in the first line, and enter the array sequence in the second line.

For 50% of the data: 1<=n<=10000;

For 100% data: 1<=n<=500000;

Output description:

The calculated maximum value of the output array.

Code:

import java.util.Scanner;

public class MaxRange {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int arr[] = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = in.nextInt();
		}
		in.close();
		System.out.println(getMax(arr, 0, n - 1));
	}

	private static int getMax(int[] arr, int start, int end) {
		if (arr == null || start > end) {
			return 0;
		}
		int n = end - start + 1;
		int[][] min = new int[n + 1][n + 1];
		int[] sum = new int[n + 1];
		sum[0] = 0;
		// sum[i]即从第一个数加到第i个数的和，也就是arr[0]+...+arr[i-1]
		for (int i = start + 1; i <= end + 1; i++) {
			sum[i - start] = sum[i - start - 1] + arr[i - start - 1];
		}

		int max = -1;
		for (int k = 0; k <= end - start; k++)
			// 左右下标的差，k==0时,区间内有1个数
			for (int i = 0; i <= end - start - k; i++) {
				int j = i + k;
				if (k == 0) {
					min[i][j] = arr[i];
				} else {
					if (arr[j] < min[i][j - 1]) {
						min[i][j] = arr[j];
					} else {
						min[i][j] = min[i][j - 1];
					}
				}
				max = Math.max(max, min[i][j] * (sum[j + 1] - sum[i]));
			}

		return max;
	}
}