Question

我试图让节目接近The Maximum Volume of Trapped Rain Water in 3D。该计划必须输出下雨后捕获的水量。假设输入是[0,1,0,2,1,0,1,3,2,1,2,1]

http://i.stack.imgur.com/mbX7I.png

输出应为6

我已经在我的代码中实现了他的算法，并且我在评论中提到了1个错误，说明＆＃34;无法解决方法max（int，double）＆＃34;。

任何人都可以解释我如何解决它？我对java很新。谢谢。

import java.util.Scanner;
import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        Scanner scn = new Scanner(System.in);
        int[] numbers = new int[10];
        Main obj = new Main();

        System.out.println("Enter 10 numbers");
        for (int i = 0; i < numbers.length; i++) {
            numbers[i] = scn.nextInt();
        }
        System.out.println("Numbers entered: ");
        obj.output(numbers);

        int min = Arrays.stream(numbers).min().getAsInt();
        int max = Arrays.stream(numbers).max().getAsInt();
        System.out.println("Min = " + min);
        System.out.println("Max = " + max);


        System.out.println("Second max = " + obj.secondHighest(numbers));

    }

    public void output(int[] numbers) {
        for (int i = 0; i < numbers.length; i++) {
            System.out.println(numbers[i]);
        }

    }

    public int secondHighest(int[] numbers) {
        int high1 = Integer.MIN_VALUE;
        int high2 = Integer.MIN_VALUE;
        for (int num : numbers) {
            if (num > high1) {
                high2 = high1;
                high1 = num;
            } else if (num > high2) {
                high2 = num;
            }
        }
        return high2;
    }

    int trapWaterVolume(int[] numbers) {
        int n = numbers.length;
        int[] leftmost = new int[n];
        int[] rightmost = new int[n];
        int leftMaxSoFar = 0;
        for (int i = 0; i < n; i++) {
            leftmost[i] = leftMaxSoFar;
            if (numbers[i] > leftMaxSoFar) leftMaxSoFar = numbers[i];
        }

        //right exclusive scan, O(n), the highest bar to the right each point
        int rightMaxSoFar = 0;
        for (int i = n - 1; i >= 0; i--){
            rightmost[i] = rightMaxSoFar;
            if (numbers[i] > rightMaxSoFar) rightMaxSoFar = numbers[i];
        }


        // Summation, O(n)
        int vol = 0;
        for (int i = 0; i < n; i++){
            vol += max(0, min(leftmost[i], (rightmost[i]) - (numbers[i]))); //Error @max
            // can't resolve method max(int, double)
        }
        return vol;


    }


}

Answer 1

使用Math.max代替max，因为您尚未声明max方法，但Math类中已有一个。

vol += Math.max(0, min(leftmost[i], (rightmost[i]) - (numbers[i])));

被困在2D模型Java

1 个答案: