Rectangle Area

最新推荐文章于 2018-12-03 20:09:00 发布

iteye_6051

最新推荐文章于 2018-12-03 20:09:00 发布

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分类专栏： Leetcode 文章标签： Math

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本文介绍了一种方法来计算两个矩形在二维平面内的总面积，并且通过减去它们的重叠区域来得出总面积差。通过提供代码示例，详细解释了计算过程。

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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：


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);
    }
}