223. Rectangle Area

class Solution(object):
    def computeArea(self, A, B, C, D, E, F, G, H):
        """
        :type A: int
        :type B: int
        :type C: int
        :type D: int
        :type E: int
        :type F: int
        :type G: int
        :type H: int
        :rtype: int
        """
        overlap = max(min(C,G)-max(A,E), 0)*max(min(D,H)-max(B,F), 0)

        return (A-C)*(B-D) + (E-G)*(F-H) - overlap

http://bookshadow.com/weblog/2015/06/08/leetcode-rectangle-area/

<h2>题目大意：</h2>   <p>计算二维平面上两个直线矩形的覆盖面积。</p>   <p>矩形通过其左下角和右上角的坐标进行定义。</p>   <p>假设总面积不会超过int的最大值。</p>   <h2>解题思路：</h2>   <p>简单计算几何。根据容斥原理：S(M &cup; N) = S(M) + S(N)&nbsp;- S(M &cap; N)</p>   <p>题目可以转化为计算矩形相交部分的面积</p>   <p>S(M) =&nbsp;(C - A) * (D - B)</p>   <p>S(N) =&nbsp;(G - E) * (H - F)</p>   <p>S(M &cap; N) =&nbsp;max(min(C, G) - max(A, E), 0) * max(min(D, H) - max(B, F), 0)</p>