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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：问题的关键点是判断是否两个矩形存在覆盖。
若存在覆盖时，必定有：A和E的最大值要小于C和G的最小值；B和F的最大值要小于D和H的最小值。
因此我们有如下代码
1 class Solution { 2 public: 3 int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) { 4 int left = max(A, E), right = max(min(C, G), left); 5 int bot = max(B, F), top = max(min(D, H), bot); 6 return (C - A) * (D - B) + (G - E) * (H - F) 7 - (right - left) * (top - bot); 8 } 9 };
其中减去的部分就是重叠的面积，当不重叠时该值为0。
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原文地址：http://www.cnblogs.com/fenshen371/p/4908155.html
