LeetCode/src/main/java/L0223_RectangleArea.java at master · LjyYano/LeetCode · GitHub

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/**

* https://leetcode.cn/problems/rectangle-area/

* 在二维平面上实现一个函数，计算其中两个由直线构成的矩形重叠后形成的总面积。

* 每个矩形由其左下顶点和右上顶点坐标表示。

* 示例 1：

* 输入：ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2

* 输出：45

* 示例 2：

* 输入：ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2

* 输出：16

* 提示：

* -10^4 <= ax1, ay1, ax2, ay2, bx1, by1, bx2, by2 <= 10^4

public class L0223_RectangleArea {

public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {

// 计算两个矩形的面积

int area1 = (ax2 - ax1) * (ay2 - ay1);

int area2 = (bx2 - bx1) * (by2 - by1);

// 计算重叠区域的坐标

int overlapX1 = Math.max(ax1, bx1);

int overlapY1 = Math.max(ay1, by1);

int overlapX2 = Math.min(ax2, bx2);

int overlapY2 = Math.min(ay2, by2);

// 计算重叠区域的面积

int overlapArea = 0;

if (overlapX2 > overlapX1 && overlapY2 > overlapY1) {

overlapArea = (overlapX2 - overlapX1) * (overlapY2 - overlapY1);

}

// 返回总面积 = 两个矩形的面积和 - 重叠区域的面积

return area1 + area2 - overlapArea;

}

public static void main(String[] args) {

L0223_RectangleArea solution = new L0223_RectangleArea();

// 测试用例 1

System.out.println(solution.computeArea(-3, 0, 3, 4, 0, -1, 9, 2)); // 预期输出：45

// 测试用例 2

System.out.println(solution.computeArea(-2, -2, 2, 2, -2, -2, 2, 2)); // 预期输出：16

}

LeetCode/src/main/java/L0223_RectangleArea.java at master · LjyYano/LeetCode · GitHub

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