391. Perfect Rectangle

矩形区域覆盖判断算法

最新推荐文章于 2025-07-26 10:42:35 发布

Alden He

最新推荐文章于 2025-07-26 10:42:35 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/u014683187/article/details/90034421

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本文介绍了一种算法，用于判断一组轴对齐的矩形是否完全覆盖一个矩形区域，且无重叠和空隙。通过遍历矩形集合，确定最大轮廓，并使用哈希集记录顶点出现次数，最终验证覆盖的有效性。

Given N axis-aligned rectangles where N > 0, determine if they all together form an exact cover of a rectangular region.

Each rectangle is represented as a bottom-left point and a top-right point. For example, a unit square is represented as [1,1,2,2]. (coordinate of bottom-left point is (1, 1) and top-right point is (2, 2)).

Example 1:

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

Return true. All 5 rectangles together form an exact cover of a rectangular region.

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Example 2:

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

Return false. Because there is a gap between the two rectangular regions.

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Example 3:

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

Return false. Because there is a gap in the top center.

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Example 4:

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

Return false. Because two of the rectangles overlap with each other.

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思路：遍历一遍所有的矩形，得到所有矩形的最大轮廓，在遍历过程中记录矩形的四个顶点构成的坐标出现的次数，如果是完美矩形，则这些顶点出现的次数出轮廓顶点外会出现偶数次，轮廓顶点出现一次

public boolean isRectangleCover(int[][] rectangles) {
       int top=rectangles[0][2];
       int right=rectangles[0][3];
       int left=rectangles[0][1];
       int bottom=rectangles[0][0];
       HashSet<String> set = new HashSet<String>();
       int area=0;
       for(int[] rectangle:rectangles){
           top=Math.max(top,rectangle[2]);
           right=Math.max(right,rectangle[3]);
           left=Math.min(left,rectangle[1]);
           bottom=Math.min(bottom,rectangle[0]);
           String s1 = rectangle[0] + " " + rectangle[1];
           String s2 = rectangle[0] + " " + rectangle[3];
           String s3 = rectangle[2] + " " + rectangle[3];
           String s4 = rectangle[2] + " " + rectangle[1];
           area+=(rectangle[2]-rectangle[0])*(rectangle[3]-rectangle[1]);
           if (!set.add(s1)) set.remove(s1);
           if (!set.add(s2)) set.remove(s2);
           if (!set.add(s3)) set.remove(s3);
           if (!set.add(s4)) set.remove(s4);

       }
       if (!set.contains(bottom + " " + left) || !set.contains(bottom + " " + right) || !set.contains(top + " " + left) || !set.contains(top + " " + right) || set.size() != 4) return false;
       return area==(top-bottom)*(right-left);
   }