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.
求矩形的面积,可以分为三种情况:两个矩形左右分离;上下分离;相交;
相交时求左边下边的最大值,右边上边的最小值。
public int computeArea(int A,int B,int C,int D,int E,int F,int G,int H) {
if(A>G||E>C)
return (C-A)*(D-B)+(G-E)*(H-F);
if(B>H||F>D)
return (C-A)*(D-B)+(G-E)*(H-F);
int left=Math.max(A, E);
int right=Math.min(C, G);
int top=Math.min(D, H);
int bottom=Math.max(B, F);
return (C-A)*(D-B)+(G-E)*(H-F)-(right-left)*(top-bottom);
}