LeetCode 836. 矩形重叠

注：坐标系y轴是向上的

自己想到的

	public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        int x1 = rec1[0], x2 = rec1[2], x3 = rec2[0], x4 = rec2[2];
        int y1 = rec1[1], y2 = rec1[3], y3 = rec2[1], y4 = rec2[3];

        boolean one = (x2>x3 && y2>y3) && (x1<x4 && y1<y4); //第一种情况rec1在左下，rec2在右上
        boolean two = (x2<x3 && y2<y3) && (x1>x4 && y1>y4); //反过来

        return one || two;
    }

官方题解一

  public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        return !(rec1[2] <= rec2[0] ||   // left
                 rec1[3] <= rec2[1] ||   // bottom
                 rec1[0] >= rec2[2] ||   // right
                 rec1[1] >= rec2[3]);    // top
    }

考虑不重叠的情况，最后取反

官方题解二

 public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        return (Math.min(rec1[2], rec2[2]) > Math.max(rec1[0], rec2[0]) &&
                Math.min(rec1[3], rec2[3]) > Math.max(rec1[1], rec2[1]));
    }

• 重叠区域也是一个矩形，则两个矩形的水平边投影到 x 轴上时会有交集，y轴同理也会有交集