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.
Rectangle Area 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.
这道题很简单，关键就是要确定如何求两个矩形如果存在重复，重复区域的宽和高怎么确定。
可以这么来确定重复矩形的宽（高同理）：用两个矩形中右边的最小值减去左边的最大值，如果这个值大于0，表示存在重复，如果这个值小于0，，表示不存在重复。 runtime:0ms
class Solution {public:int computeArea, , , ) {int area=(C-A)*(D-B)+(G-E)*(H-F);int width=max(min(C,G)-max(A,E),0);int height=max(min(D,H)-max(B,F),0);return area-width*height;}};
快乐要懂得分享，才能加倍的快乐
