223. Rectangle Area

Find the total area covered by two rectilinear rectangles in a2D 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.

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给出两个矩形的两个对角的坐标（也就是给两个矩形），要求求出总面积，注意会有重叠。求面积比较简单，比如第一个矩形的面积就是(C - A) * (D - B)。关键部分就是求重叠部分的面积。首先要直到是否重叠，先找出A和E中比较大的设为l，再找出C和G中比较小的设为r，自己画图画出几种情况就能发现l小于r时才能重叠，这是左右的关系；同理上下的关系也类似，找出u和d再比较。如果有重叠，重叠的面积为(r - l) * (u - d)。最后用两个矩形的面积的和减去重叠面积即可得到答案。

代码：

class Solution{public:int computeArea(int A, int B, int C, int D, int E, int F, int G, int H){int intersection = 0;int l = max(A, E), r = min(C, G), d = max(B, F), u = min(D, H);if(l < r && d < u){intersection = (r - l) * (u - d);}return (C - A) * (D - B) + (G - E) * (H - F) - intersection; }};