[LeetCode]60. Rectangle Area矩形面积

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.

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解法：注意前提条件：矩形的边和坐标轴是平行的。所有的不相交的情况只有四种：一个矩形在另一个的上下左右四个位置不重叠，这四种情况下返回两个矩形面积之和。其他所有情况下两个矩形是有交集的。由于交集都是在中间，所以横边的左端点是两个矩形左顶点横坐标的较大值，右端点是两个矩形右顶点的较小值，同理，竖边的下端点是两个矩形下顶点纵坐标的较大值，上端点是两个矩形上顶点纵坐标的较小值。

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int res = (C - A) * (D - B) + (G - E) * (H - F);
        if (!(A >= G || B >= H || E >= C || F >= D))
            res -= (min(G, C) - max(E, A)) * (min(H, D) - max(F, B));
        return res;
    }
};