[Leetcode] 223. Rectangle Area 解题报告_total area covered by rectangles-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/73649633

本文介绍了一种计算二维平面上两个直角矩形覆盖总面积的方法。通过判断矩形是否重叠及重叠区域的大小来计算总面积，并提供了两种简洁的C++实现代码。

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题目：

Find the total area covered by two rectilinear rectangles in a 2D plane.

Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.

Assume that the total area is never beyond the maximum possible value of int.

思路：

一道简单的数学题，关键在于计算出来重合部分的面积，这里重点在于区分各种重合情况。我自己写的代码比较冗长，这里同时贴几个写的很精炼的代码。

代码：

1、自己的代码：

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int area1 = (C - A) * (D - B);
    	int area2 = (G - E) * (H - F);
    	if (E >= C || G <= A || H <= B || F >= D) {     // no overlap
    		return area1 + area2;
    	}
    	int width = 0, height = 0;
    	if (E < A) {                                    // update width
    		width = G < C ? (G - A) : (C - A);
    	}
    	else {
    		width = G < C ? (G - E) : (C - E);
    	}
    	if (F < B) {                                    // update height
    		height = H < D ? (H - B) : (D - B);
    	}
    	else {
    		height = H < D ? (H - F) : (D - F);
    	}
    	return area1 + area2 - width * height;
    }
};

2、别人的代码：

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int area = (C - A) * (D - B) + (G - E) * (H - F);  
        if(!(B >= H || F >= D || A >= G ||E >= C)) {  
            int tem = (min(C, G) - max(A, E)) * (min(D, H) -max(B, F));  
            area -= tem;  
        }  
        return area;  
    }
};

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int left = max(A, E), right = max(min(C, G), left);  
        int bot = max(B, F), top = max(min(D, H), bot);  
        return (C - A) * (D - B) + (G - E) * (H - F) - (right - left) * (top - bot);
    }
};