The two functions below will do the job quickly.

In both cases the function parameters represent the top-left and bottom-right coordinates for the 2 rectangles. The first method returns true if the rectangles overlap. The second  method returns an array containing the top-left and bottom-right coordinates of any overlap region, if there is no overlap it returns an array of length zero.

Function 1

1. boolean box_box(float ax0, float ay0, float ax1, float ay1, float bx0, float by0, float bx1, float by1){
2.     float topA = (float) Math.min(ay0, ay1);
3.     float botA = (float) Math.max(ay0, ay1);
4.     float leftA = (float) Math.min(ax0, ax1);
5.     float rightA = (float) Math.max(ax0, ax1);
6.     float topB = (float) Math.min(by0, by1);
7.     float botB = (float) Math.max(by0, by1);
8.     float leftB = (float) Math.min(bx0, bx1);
9.     float rightB = (float) Math.max(bx0, bx1);
10.     if(botA <= topB  || botB <= topA || rightA <= leftB || rightB <= leftA)
11.         return false;
12.     return true;   
13. }
14.    
15. float[] box_box_p(float ax0, float ay0, float ax1, float ay1, float bx0, float by0, float bx1, float by1){
16.     float[] result = new float[0];
17.     float topA = (float) Math.min(ay0, ay1);
18.     float botA = (float) Math.max(ay0, ay1);
19.     float leftA = (float) Math.min(ax0, ax1);
20.     float rightA = (float) Math.max(ax0, ax1);
21.     float topB = (float) Math.min(by0, by1);
22.     float botB = (float) Math.max(by0, by1);
23.     float leftB = (float) Math.min(bx0, bx1);
24.     float rightB = (float) Math.max(bx0, bx1);
25.    
26.     if(botA <= topB  || botB <= topA || rightA <= leftB || rightB <= leftA)
27.         return result;
28.     float leftO = (leftA < leftB) ? leftB : leftA;
29.     float rightO = (rightA > rightB) ? rightB : rightA;
30.     float botO = (botA > botB) ? botB : botA;
31.     float topO = (topA < topB) ? topB : topA;
32.     result =  new float[] {leftO, topO, rightO, botO};
33.     return result;
34. }