Leet Code OJ 223. Rectangle Area [Difficulty: Easy]

题目：
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.

翻译：
在二维平面中，计算被2个由直线围成的矩形所覆盖的总面积。
每个矩形都是由它左下角和右上角的点来定义的。

分析：
这道题目的关键是需要画图理清可能出现的情况。
1. 2个矩形不相交，这边分为2种情况，一个是第二个矩形在第一个矩形的上面或者下面；一个是第二个矩形在第一个矩形的左边或者右边；也可能2种都有，但是由于是或的关系，所以不用考虑。
2. 2个矩形相交，相交其实就是把4个x轴放在一起，找出其中大小在中间的2个，作为x轴；y轴同理。

代码：

public class Solution {public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {int sum=(C-A)*(D-B)+(G-E)*(H-F);if(C<E||G<A||H<B||D<F){//没有交叉return sum;}else{//有交叉//计算重叠部分的x坐标，ACEGint x1=A<E?E:A;int x2=C<G?C:G;//计算重叠部分的y坐标, BDFHint y1=B<F?F:B;int y2=D<H?D:H;return sum-(x2-x1)*(y2-y1);}}
}

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