Question

223. 矩形面积

English Version

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

解法

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 $O(1)$。

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)

class Solution {
  public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
    int a = (ax2 - ax1) * (ay2 - ay1);
    int b = (bx2 - bx1) * (by2 - by1);
    int width = Math.min(ax2, bx2) - Math.max(ax1, bx1);
    int height = Math.min(ay2, by2) - Math.max(ay1, by1);
    return a + b - Math.max(height, 0) * Math.max(width, 0);
  }
}

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

func computeArea(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)
}

function computeArea(
  ax1: number,
  ay1: number,
  ax2: number,
  ay2: number,
  bx1: number,
  by1: number,
  bx2: number,
  by2: number,
): number {
  const a = (ax2 - ax1) * (ay2 - ay1);
  const b = (bx2 - bx1) * (by2 - by1);
  const width = Math.min(ax2, bx2) - Math.max(ax1, bx1);
  const height = Math.min(ay2, by2) - Math.max(ay1, by1);
  return a + b - Math.max(width, 0) * Math.max(height, 0);
}

public class Solution {
  public int ComputeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
    int a = (ax2 - ax1) * (ay2 - ay1);
    int b = (bx2 - bx1) * (by2 - by1);
    int width = Math.Min(ax2, bx2) - Math.Max(ax1, bx1);
    int height = Math.Min(ay2, by2) - Math.Max(ay1, by1);
    return a + b - Math.Max(height, 0) * Math.Max(width, 0);
  }
}