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](https://leetcode.com/static/images/problemset/rectangle_area.png)
Assume that the total area is never beyond the maximum possible value of int.
public class Solution { public static void main(String args[]){ Solution a = new Solution(); int c =a.computeArea(-2,-2,2,2,3,3,4,4); System.out.println(c); } public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) { int sqr = 0; sqr = (C-A>0? (C-A):(A-C) )* (D-B>0?(D-B):(B-D)) +(H>F? H-F:F-H)*(G>E?G-E:E-G); int result ; result = sqr - area(min(C,G),min(D,H),max(A,E),max(B,F)); return(E>C||F>D||B>H||A>G)? sqr:result; } public int area(int A, int B, int C, int D){ int sqr; sqr = (C-A>0? (C-A):(A-C) )* (D-B>0?(D-B):(B-D)) ; return sqr; } public int min(int a,int b){ int c; c = a>b? b:a; return c; } public int max(int a,int b){ int c; c = a>b? a:b; return c; } }
//没考虑清楚,求面积不用判断符号,没有啥捷径目前发现
分析各种情况,总结条件,编写代码
有点像设计数字电路