Leetcode #223 Rectangle Area

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.

Credits:

Special thanks to
@mithmatt for adding this problem, creating the above image and all test cases.

class Solution {
public:
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
int ans = (C- A) * (D - B) + (G - E) * (H - F);
if(C <= E || G <= A || D <= F || H <= B) return ans;  //两矩形不相交
int x[4] = {A, C, E, G};   int y[4] = {B, D, F, H};
sort(x, x + 4); sort(y, y + 4);
return ans - (x[2] - x[1]) * (y[2] - y[1]);          //分别取在x，y轴方向上相交线段的长度相乘即为重合部分面积。
}
};

对两矩形是否相交的判断及处理可以进一步简化，最终代码只需一行：

return (C-A) * (D-B) + (G-E) * (H-F) - max(0, min(G,C) - max(A,E)) * max(0, min(D,H) - max(B,F));

但提交后却在最后一组数据WA了。



原因在于该组case中 计算min(G, C) - max(A,E)时数据溢出。

故仅需修改成long long型，如下，即可顺利AC。

return (C-A) * (D-B) + (G-E) * (H-F) - max((long long)0, ((long long)min(G,C) - (long long) max(A,E))) * max((long long)0,  (long long)(min(D,H) - max(B,F)));