391. Perfect Rectangle

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.

 

 

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.

 

 

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.

 

 

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.

这个题我甚至不知道该怎么总结。

难就难在从这个题抽象出一种解法，看了别人的答案和思路= =然而没有归类总结到某种类型，这题相当于背了个题。。。

简单的说，除了最外面的4个点，所有的点都会2次2次的出现，如果有覆盖，覆盖进去的点就不是成对出现了。

最外面4个点围的面积等于所有小矩形面积的和。

就用这2个判断就行了。

判断成对的点用的SET，单次出现添加，第二次出现删除。。这样最后应该都删掉，SET里只剩下4个最外面的点。

剩下的就是判断最外点，不停地更新。。

Consider how the corners of all rectangles appear in the large rectangle if there's a perfect rectangular cover.
Rule1: The local shape of the corner has to follow one of the three following patterns

• Corner of the large rectangle (blue): it occurs only once among all rectangles
• T-junctions (green): it occurs twice among all rectangles
• Cross (red): it occurs four times among all rectangles

For each point being a corner of any rectangle, it should appear even times except the 4 corners of the large rectangle. So we can put those points into a hash map and remove them if they appear one more time.

At the end, we should only get 4 points. 

Rule2:  the large rectangle area should be equal to the sum of small rectangles

public class Solution {
    public boolean isRectangleCover(int[][] rectangles) {
        if (rectangles==null || rectangles.length==0 || rectangles[0].length==0) return false;
        int subrecAreaSum = 0;  //sum of subrectangle's area
        int x1 = Integer.MAX_VALUE; //large rectangle bottom left x-axis
        int y1 = Integer.MAX_VALUE; //large rectangle bottom left y-axis
        int x2 = Integer.MIN_VALUE; //large rectangle top right x-axis
        int y2 = Integer.MIN_VALUE; //large rectangle top right y-axis
        
        HashSet<String> set = new HashSet<String>(); // store points
        
        for(int[] rec : rectangles) {
            //check if it has large rectangle's 4 points
            x1 = Math.min(x1, rec[0]);
            y1 = Math.min(y1, rec[1]);
            x2 = Math.max(x2, rec[2]);
            y2 = Math.max(y2, rec[3]);
            
            //calculate sum of subrectangles
            subrecAreaSum += (rec[2]-rec[0]) * (rec[3] - rec[1]);
            
            //store this rectangle's 4 points into hashSet
            String p1 = Integer.toString(rec[0]) + "" + Integer.toString(rec[1]);
            String p2 = Integer.toString(rec[0]) + "" + Integer.toString(rec[3]);
            String p3 = Integer.toString(rec[2]) + "" + Integer.toString(rec[1]);
            String p4 = Integer.toString(rec[2]) + "" + Integer.toString(rec[3]);
            
            if (!set.add(p1)) set.remove(p1);
            if (!set.add(p2)) set.remove(p2);
            if (!set.add(p3)) set.remove(p3);
            if (!set.add(p4)) set.remove(p4);
        }
        
        if (set.size()!=4 || !set.contains(x1+""+y1) || !set.contains(x1+""+y2) || !set.contains(x2+""+y1) || !set.contains(x2+""+y2))
            return false;
        return subrecAreaSum == (x2-x1) * (y2-y1);
    }
}

　矩阵的题不光会heap, bfs,dfs, 还有单纯的找规律, 遍历四周, 01编码, 求和, map存边角为字符串

存为字符串 //store this rectangle's 4 points into hashSet
            String p1 = Integer.toString(rec[0]) + "" + Integer.toString(rec[1]);

boolean

add(E e)            如果此 set 中尚未包含指定元素，则添加指定元素。

如果此 set 已经包含该元素，则该调用不改变此 set 并返回 false。结合构造方法上的限制，这就可以确保 set 永远不包含重复的元素