*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.

[LeetCode]Rectangle Area

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

题目大意：

计算二维平面上两个直线矩形的覆盖面积。

矩形通过其左下角和右上角的坐标进行定义。

假设总面积不会超过int的最大值。

解题思路：

简单计算几何。根据容斥原理：S(M ∪ N) = S(M) + S(N) - S(M ∩ N)

题目可以转化为计算矩形相交部分的面积

S(M) = (C - A) * (D - B)

S(N) = (G - E) * (H - F)

S(M ∩ N) = max(min(C, G) - max(A, E), 0) * max(min(D, H) - max(B, F), 0)

If they do not have overlap, the total area is the sum of 2 rectangle areas. If they have overlap, the total area should minus the overlap area.

 1 public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
 2     if(C<E||G<A )
 3         return (G-E)*(H-F) + (C-A)*(D-B);
 4  
 5     if(D<F || H<B)
 6         return (G-E)*(H-F) + (C-A)*(D-B);
 7  
 8     int right = Math.min(C,G);
 9     int left = Math.max(A,E);
10     int top = Math.min(H,D);
11     int bottom = Math.max(F,B);
12  
13     return (G-E)*(H-F) + (C-A)*(D-B) - (right-left)*(top-bottom);
14 }

reference: http://www.programcreek.com/2014/06/leetcode-rectangle-area-java/

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