3000. Maximum Area of Longest Diagonal Rectangle

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# 題目敘述

You are given a 2D 0-indexed integer array dimensions .

For all indices i , 0 <= i < dimensions.length , dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i .

Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area.

# Example 1

Input: dimensions = [[9,3],[8,6]]
Output: 48
Explanation:
For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.
For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.

# Example 2

Input: dimensions = [[3,4],[4,3]]
Output: 12
Explanation: Length of diagonal is the same for both which is 5, so maximum area = 12.

# 解題思路

基本上按照題目邏輯完成函式就可以。

# Solution

class Solution {

    public int areaOfMaxDiagonal(int[][] dimensions) {

        int result = 0;

        double max_area = 0;

        for (int i = 0; i < dimensions.length; i++) {

            double area = dimensions[i][0] * dimensions[i][0] + dimensions[i][1] * dimensions[i][1];

            area = Math.sqrt(area);

            if (area > max_area) {

                max_area = area;

                result = dimensions[i][0] * dimensions[i][1];

            }else if (area == max_area) {

                result = Math.max(result, dimensions[i][0] * dimensions[i][1]);

            }

        }

        return result;

    }

}

#include <vector>

#include <cmath>

#include <algorithm>

class Solution {

public:

    int areaOfMaxDiagonal(std::vector<std::vector<int>>& dimensions) {

        int max_area = 0;

        int result = 0;

        for (int i = 0; i < dimensions.size(); i++) {

            double area = sqrt(pow(dimensions[i][0], 2) + pow(dimensions[i][1], 2));

            if (area > max_area) {

                max_area = area;

                result = dimensions[i][0] * dimensions[i][1];

            } else if (area == max_area) {

                result = std::max(result, dimensions[i][0] * dimensions[i][1]);

            }

        }

        return result;

    }

};