题目描述
给你 二维 平面上两个 由直线构成的 矩形，请你计算并返回两个矩形覆盖的总面积。 每个矩形由其 左下 顶点和 右上 顶点坐标表示： 第一个矩形由其左下顶点 (ax1, ay1) 和右上顶点 (ax2, ay2) 定义。 第二个矩形由其左下顶点 (bx1, by1) 和右上顶点 (bx2, by2) 定义。
Example1:
输入：ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2 输出：45
Example2:
输入：ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2 输出：16
Note:
- -10^4 <= ax1, ay1, ax2, ay2, bx1, by1, bx2, by2 <= 10^4
解题思路
- 需要找到矩形重叠的坐标
- 注意无重叠的情况
class Solution {
public:
int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
int area1 = (ax2 - ax1) * (ay2 - ay1);
int area2 = (bx2 - bx1) * (by2 - by1);
int overlapWidth = min(ax2, bx2) - max(bx1, ax1);
int overlapHeight = min(ay2, by2) - max(ay1, by1);
int overlapArea = max(overlapWidth, 0) * max(overlapHeight, 0);
return area1 + area2 - overlapArea;
}
};
