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.[img]https://leetcode.com/static/images/problemset/rectangle_area.png[/img]
Assume that the total area is never beyond the maximum possible value of int.
首先计算出两个长方形的面积,然后计算出它们重叠的面积,最后返回两个面积的差就可以了。代码如下:
Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.[img]https://leetcode.com/static/images/problemset/rectangle_area.png[/img]
Assume that the total area is never beyond the maximum possible value of int.
首先计算出两个长方形的面积,然后计算出它们重叠的面积,最后返回两个面积的差就可以了。代码如下:
public class Solution {
public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
int area1 = (C - A) * (D - B);
int area2 = (G - E) * (H - F);
return area1 + area2 - (int)getOverlap(A, B, C, D, E, F, G, H);
}
public int getOverlap(int A, int B, int C, int D, int E, int F, int G, int H) {
long w1 = Math.max(A, E);
long w2 = Math.min(C, G);
long w = w2 - w1;
long h1 = Math.min(D, H);
long h2 = Math.max(B, F);
long h = h1 - h2;
if(w <= 0 || h <= 0) return 0;
return (int) (w * h);
}
}