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计算两个矩形重叠的面积

时间:2014-01-20 09:04:29      阅读:468      评论:0      收藏:0      [点我收藏+]
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private int x1;
    private int y1;
    private int width1;
    private int height1;
    
    private int x2;
    private int y2;
    private int width2;
    private int height2;
    
    private int endx;
    private int startx;
    private int endy;
    private int starty;
    
    private int height;
    private int width;
    
    private int area;
    private int area1;
    private int area2;
    
    private float ratio;
    
    /**
     * 检测两个矩形是否重叠,重叠的面积为50以上才认定为重叠
     */
    private boolean isOverlap(View view1, View view2) {
        x1 = view1.getLeft();
        y1 = view1.getRight();
        width1 = view1.getWidth();
        height1 = view1.getHeight();

        x2 = view2.getLeft();
        y2 = view2.getRight();
        width2 = view2.getWidth();
        height2 = view2.getHeight();

        endx = Math.max(x1 + width1, x2 + height2);
        startx = Math.min(x1, x2);
        width = width1 + width2 - (endx - startx);

        endy = Math.max(y1 + height1, y2 + height2);
        starty = Math.min(y1, y2);
        height = height1 + height2 - (endy - starty);

        if (width <= 0 || height <= 0)
            return false;
        else {
            area = width * height;
            area1 = width1 * height1;
            area2 = width2 * height2;
            ratio = area / (area1 + area2 - area);
            if (ratio >= 0.5) {
                return true;
            }
        }
        return false;
    }
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进行了一万次的运算用了45毫秒

计算两个矩形重叠的面积

原文:http://www.cnblogs.com/622698abc/p/3526387.html

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