223. 矩形面积

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。
 

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:

  • -104 <= ax1, ay1, ax2, ay2, bx1, by1, bx2, by2 <= 104

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/rectangle-area

 

枚举

class Solution:
    def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
        if bx1<=ax1:
            ax1,bx1=bx1,ax1
            ay1,by1=by1,ay1
            ax2,bx2=bx2,ax2
            ay2,by2=by2,ay2
        s=(ax2-ax1)*(ay2-ay1)+(bx2-bx1)*(by2-by1)
        if by1>=ay2 or by2<=ay1 or bx1>=ax2:d=0
        elif bx2>ax2:
            if by2<=ay2:d=(by2-ay1)*(ax2-bx1)
            if by1>=ay1:d=(ay2-by1)*(ax2-bx1)
            if ay2<=by2 and ay1>=by1:d=(ay2-ay1)*(ax2-bx1)
            if by2<=ay2 and by1>=ay1:d=(by2-by1)*(ax2-bx1)
        else:
            if by2<=ay2:d=(by2-ay1)*(bx2-bx1)
            if by1>=ay1:d=(ay2-by1)*(bx2-bx1)
            if ay2<=by2 and ay1>=by1:d=(ay2-ay1)*(bx2-bx1)
            if by2<=ay2 and by1>=ay1:d=(by2-by1)*(bx2-bx1)
        return s-d
            

数学方法

class Solution:
    def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
        ax, bx = ax2 - ax1, bx2 - bx1
        ay, by = ay2 - ay1, by2 - by1
        cx = max(ax1, ax2, bx1, bx2) - min(ax1, ax2, bx1, bx2)
        cy = max(ay1, ay2, by1, by2) - min(ay1, ay2, by1, by2)
        x = ax + bx - cx
        y = ay + by - cy
 
        if ax2 <= bx1 or bx2 <= ax1 or ay2 <= by1 or by2 <= ay1:
            return ax * ay + bx * by

        return ax * ay + bx * by - x * y

 

posted @ 2021-09-30 11:57  XXXSANS  阅读(185)  评论(0编辑  收藏  举报