0223. 矩形面积【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - 数学

1. 📝 题目描述

• leetcode

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。

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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

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提示：

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

2. 🎯 s.1 - 数学

c

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 overlapW = fmax(0, fmin(ax2, bx2) - fmax(ax1, bx1));
    int overlapH = fmax(0, fmin(ay2, by2) - fmax(ay1, by1));
    return area1 + area2 - overlapW * overlapH;
}

js

/**
 * @param {number} ax1
 * @param {number} ay1
 * @param {number} ax2
 * @param {number} ay2
 * @param {number} bx1
 * @param {number} by1
 * @param {number} bx2
 * @param {number} by2
 * @return {number}
 */
var computeArea = function (ax1, ay1, ax2, ay2, bx1, by1, bx2, by2) {
  const area1 = (ax2 - ax1) * (ay2 - ay1)
  const area2 = (bx2 - bx1) * (by2 - by1)
  const overlapW = Math.max(0, Math.min(ax2, bx2) - Math.max(ax1, bx1))
  const overlapH = Math.max(0, Math.min(ay2, by2) - Math.max(ay1, by1))
  return area1 + area2 - overlapW * overlapH
}

py

class Solution:
    def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
        area1 = (ax2 - ax1) * (ay2 - ay1)
        area2 = (bx2 - bx1) * (by2 - by1)
        overlap_w = max(0, min(ax2, bx2) - max(ax1, bx1))
        overlap_h = max(0, min(ay2, by2) - max(ay1, by1))
        return area1 + area2 - overlap_w * overlap_h

• 时间复杂度：$O(1)$
• 空间复杂度：$O(1)$

算法思路：

• 总面积 = 矩形1面积 + 矩形2面积 - 重叠面积
• 重叠宽度 = $max(0,min(ax2,bx2)-max(ax1,bx1))$
• 重叠高度 = $max(0,min(ay2,by2)-max(ay1,by1))$