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.

 

 

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