total area=area1+area2-overlaparea

->So we need to calculate overlap area->calcuate the boundary coordinate of overlap area.

So,I listed several overlap cases and tried to use conditional sentences to define area of each case.(How stupid I was ==

But after observation,I noticed the boundary of all cases has same properties:

1.the area's left is the rightmost of the left of the two rectangles.->Math.max(A,E)

2.the area's right is the leftmost of the right of the two rectangles.->Math.min(C,G)

3.the area's top is the bottommost of the top of the two rectangles.->Math.min(D,H)

4.the area's bottom is the topmost of the bottom of the two rectangles.->Math.max(B,F)

But how to represent the case which didn't overlap? If there is no overlap,right<=left or bottom<=top or both ->using quick return when test them.

```
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);
int areaLeft=Math.max(A,E);
int areaRight=Math.min(C,G);
if(areaRight<=areaLeft){
return sum;
}
int areaBottom=Math.max(B,F);
int areaTop=Math.min(D,H);
if(areaBottom>=areaTop){
return sum;
}
return sum-(areaRight-areaLeft)*(areaTop-areaBottom);
}
```