Medium

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：

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

解题思路

解题思路

这道题的核心思想是利用容斥原理：总覆盖面积 = 矩形A面积 + 矩形B面积 - 重叠部分面积。

关键步骤：

  1. 计算两个矩形的面积：矩形面积 = 长 × 宽，即 (x2-x1) × (y2-y1)

  2. 计算重叠矩形的面积：

    • 重叠矩形的左下角坐标：(max(ax1, bx1), max(ay1, by1))
    • 重叠矩形的右上角坐标：(min(ax2, bx2), min(ay2, by2))
    • 如果重叠矩形的左下角在右上角的右边或上边，说明两矩形不重叠，重叠面积为0

  3. 应用容斥原理：用两个矩形面积之和减去重叠面积

边界情况处理：

  • 两矩形完全不重叠：重叠面积为0
  • 两矩形完全重合：重叠面积等于矩形面积
  • 一个矩形包含另一个：重叠面积等于较小矩形面积

算法时间复杂度为O(1)，空间复杂度为O(1)，是最优解法。

代码实现

class Solution {
public:
    int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        // 计算两个矩形的面积
        int areaA = (ax2 - ax1) * (ay2 - ay1);
        int areaB = (bx2 - bx1) * (by2 - by1);
        
        // 计算重叠部分的左下角和右上角坐标
        int overlapLeft = max(ax1, bx1);
        int overlapRight = min(ax2, bx2);
        int overlapBottom = max(ay1, by1);
        int overlapTop = min(ay2, by2);
        
        // 计算重叠面积
        int overlapArea = 0;
        if (overlapLeft < overlapRight && overlapBottom < overlapTop) {
            overlapArea = (overlapRight - overlapLeft) * (overlapTop - overlapBottom);
        }
        
        return areaA + areaB - overlapArea;
    }
};

class Solution:
    def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
        # 计算两个矩形的面积
        area_a = (ax2 - ax1) * (ay2 - ay1)
        area_b = (bx2 - bx1) * (by2 - by1)
        
        # 计算重叠部分的左下角和右上角坐标
        overlap_left = max(ax1, bx1)
        overlap_right = min(ax2, bx2)
        overlap_bottom = max(ay1, by1)
        overlap_top = min(ay2, by2)
        
        # 计算重叠面积
        overlap_area = 0
        if overlap_left < overlap_right and overlap_bottom < overlap_top:
            overlap_area = (overlap_right - overlap_left) * (overlap_top - overlap_bottom)
        
        return area_a + area_b - overlap_area

public class Solution {
    public int ComputeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        // 计算两个矩形的面积
        int areaA = (ax2 - ax1) * (ay2 - ay1);
        int areaB = (bx2 - bx1) * (by2 - by1);
        
        // 计算重叠部分的左下角和右上角坐标
        int overlapLeft = Math.Max(ax1, bx1);
        int overlapRight = Math.Min(ax2, bx2);
        int overlapBottom = Math.Max(ay1, by1);
        int overlapTop = Math.Min(ay2, by2);
        
        // 计算重叠面积
        int overlapArea = 0;
        if (overlapLeft < overlapRight && overlapBottom < overlapTop) {
            overlapArea = (overlapRight - overlapLeft) * (overlapTop - overlapBottom);
        }
        
        return areaA + areaB - overlapArea;
    }
}

var computeArea = function(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2) {
    // 计算两个矩形的面积
    const areaA = (ax2 - ax1) * (ay2 - ay1);
    const areaB = (bx2 - bx1) * (by2 - by1);
    
    // 计算重叠部分的左下角和右上角坐标
    const overlapLeft = Math.max(ax1, bx1);
    const overlapRight = Math.min(ax2, bx2);
    const overlapBottom = Math.max(ay1, by1);
    const overlapTop = Math.min(ay2, by2);
    
    // 计算重叠面积
    let overlapArea = 0;
    if (overlapLeft < overlapRight && overlapBottom < overlapTop) {
        overlapArea = (overlapRight - overlapLeft) * (overlapTop - overlapBottom);
    }
    
    return areaA + areaB - overlapArea;
};
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复杂度分析

复杂度类型复杂度说明
时间复杂度O(1)只需要进行常数次数学运算
空间复杂度O(1)只使用了常数个变量存储中间结果

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