# coding=utf-8
# Creator:Mr.Zhao
# Creation time:2024/3/19 21:23
# 矩形面积
"""
给你 二维 平面上两个 由直线构成且边与坐标轴平行/垂直 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。
"""
ax1 = -3
ay1 = 0
ax2 = 3
ay2 = 4
bx1 = 0
by1 = -1
bx2 = 9
by2 = 2
class Solution:
def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
sjx1 = (ax2 - ax1) * (ay2 - ay1)
sjx2 = (bx2 - bx1) * (by2 - by1)
ax = {i for i in range(ax1, ax2 + 1)}
ay = {i for i in range(ay1, ay2 + 1)}
bx = {i for i in range(bx1, bx2 + 1)}
by = {i for i in range(by1, by2 + 1)}
if len(ax & bx) - 1 < 0 or len(ay & by) - 1 < 0:
return sjx2 + sjx1
else:
return sjx2 + sjx1 - (len(ax & bx) - 1) * (len(ay & by) - 1)
"""
这道题比较简单啊,应该还可以优化
交叉面积其实就是点的交集吗
"""
a = Solution()
print(a.computeArea(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2))
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