How do I compute the total area of two rectangles in the Rectangle Area problem?

Calculate each rectangle's area, compute any overlapping area, and subtract the overlap from the sum of both rectangles.

What if the rectangles do not overlap at all?

If there is no overlap, the overlap calculation will yield zero, and the total area is simply the sum of the two rectangle areas.

Can coordinates be negative?

Yes, rectangles can have negative coordinates as long as the bottom-left corner is less than or equal to the top-right corner.

How to handle rectangles touching at edges?

Edge-touching rectangles have zero overlap. Ensure your overlap formula returns max(0, overlap) to avoid negative values.

Does this approach work for more than two rectangles?

The same pattern applies, but you must iteratively calculate overlaps for all pairs to avoid double counting.

#223

Medium

auto_awesome数学·结合·几何

LeetCode 题解工作台

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 …

数学几何

题目描述

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。

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 <= ax2 <= 10⁴
-10⁴ <= ay1 <= ay2 <= 10⁴
-10⁴ <= bx1 <= bx2 <= 10⁴
-10⁴ <= by1 <= by2 <= 10⁴

lightbulb

解题思路

方法一：计算重叠面积

我们先计算出两个矩形各自的面积，记为 $a$ 和 $b$ ，然后计算重叠的宽度 $width$ 和高度 $height$ ，那么重叠的面积为 $max(width, 0) \times max(height, 0)$ ，最后将 $a$ , $b$ 和重叠面积相减即可。

时间复杂度 $O(1)$ ，空间复杂度 $O(1)$ 。

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class Solution:
    def computeArea(
        self,
        ax1: int,
        ay1: int,
        ax2: int,
        ay2: int,
        bx1: int,
        by1: int,
        bx2: int,
        by2: int,
    ) -> int:
        a = (ax2 - ax1) * (ay2 - ay1)
        b = (bx2 - bx1) * (by2 - by1)
        width = min(ax2, bx2) - max(ax1, bx1)
        height = min(ay2, by2) - max(ay1, by1)
        return a + b - max(height, 0) * max(width, 0)

speed

复杂度分析

指标	值
时间	complexity is O(1) because only constant arithmetic operations are needed. Space complexity is O(1) since no additional data structures are used beyond a few variables for coordinates and overlap calculation.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Expect clear handling of overlapping regions.
question_mark
Look for edge cases where rectangles touch but do not overlap.
question_mark
Check that the solution avoids negative overlap or double counting.

warning

常见陷阱

外企场景

error
Forgetting to subtract the overlapping area, leading to inflated totals.
error
Incorrectly calculating overlap when rectangles do not intersect.
error
Confusing width and height computations with coordinate differences.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Computing total area for multiple rectangles, not just two.
arrow_right_alt
Handling rectangles with floating-point coordinates instead of integers.
arrow_right_alt
Finding the overlapping area only without computing total area.

help

常见问题

外企场景

继续练习

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