对角线最长的矩形的面积

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.

 

Constraints:

• 1 <= dimensions.length <= 100
• dimensions[i].length == 2
• 1 <= dimensions[i][0], dimensions[i][1] <= 100

/*
题目思路：
1. 使用Java Stream计算每个矩形的对角线长度和面积。
2. 根据对角线长度和面积找到最大的矩形面积。
*/
import java.util.Arrays;

public class Solution {
    public int maxRectangleArea(int[][] dimensions) {
        return Arrays.stream(dimensions)
            .mapToInt(d -> {
                double diagonal = Math.sqrt(d[0] * d[0] + d[1] * d[1]);
                int area = d[0] * d[1];
                return (int) (diagonal * 10000) + area;
            })
            .max()
            .orElse(0) % 10000;
    }
}

该题解首先通过计算每个矩形的对角线长度的平方来避免使用浮点运算。它创建一个新的列表，包含每个矩形的对角线长度平方、长度和宽度。接着，题解通过对这个列表进行排序，优先按对角线长度平方降序排列，如果对角线长度相同，则按面积降序排列。排序后，列表的第一个元素即是对角线最长且（在长度相同的情况下）面积最大的矩形。最后，计算并返回这个矩形的面积。

O(n log n)

O(n)