leetcode 391. Perfect Rectangle

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我们有 N 个与坐标轴对齐的矩形, 其中 N > 0, 判断它们是否能精确地覆盖一个矩形区域。

每个矩形用左下角的点和右上角的点的坐标来表示。例如， 一个单位正方形可以表示为 [1,1,2,2]。 ( 左下角的点的坐标为 (1, 1) 以及右上角的点的坐标为 (2, 2) )。




示例1

rectangles = [
  [1,1,3,3],
  [3,1,4,2],
  [3,2,4,4],
  [1,3,2,4],
  [2,3,3,4]
]

返回 true。5个矩形一起可以精确地覆盖一个矩形区域。
示例2 
rectangles = [
  [1,1,2,3],
  [1,3,2,4],
  [3,1,4,2],
  [3,2,4,4]
]

返回 false。两个矩形之间有间隔，无法覆盖成一个矩形。
示例3 
rectangles = [
  [1,1,3,3],
  [3,1,4,2],
  [1,3,2,4],
  [3,2,4,4]
]

返回 false。图形顶端留有间隔，无法覆盖成一个矩形。
示例4 
rectangles = [
  [1,1,3,3],
  [3,1,4,2],
  [1,3,2,4],
  [2,2,4,4]
]

返回 false。因为中间有相交区域，虽然形成了矩形，但不是精确覆盖。

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class Solution(object):
    def isRectangleCover(self, rectangles):
        #large rectangle area is equal to small rectangles
        #large rectangle points in small rectangles points
        if len(rectangles) == 0 or len(rectangles[0]) == 0:
            return False
        x1 = float("inf")
        y1 = float("inf")
        x2 = 0
        y2 = 0
        rec = set()
        area = 0
        for points in rectangles:
            x1 = min(points[0],x1)
            y1 = min(points[1],y1)
            x2 = max(points[2],x2)
            y2 = max(points[3],y2)
            area += (points[3]-points[1])*(points[2]-points[0])
            rec.remove((points[0],points[3])) if (points[0],points[3]) in rec else rec.add((points[0],points[3]))
            rec.remove((points[0],points[1])) if (points[0],points[1]) in rec else rec.add((points[0],points[1]))
            rec.remove((points[2],points[3])) if (points[2],points[3]) in rec else rec.add((points[2],points[3]))
            rec.remove((points[2],points[1])) if (points[2],points[1]) in rec else rec.add((points[2],points[1]))
        if (x1,y2) not in rec or (x2,y1) not in rec or (x1,y1) not in rec or (x2,y2) not in rec or len(rec) != 4:
            return False
        return area == (y2-y1)*(x2-x1)