判断使用给定的坐标可否组成正方形 Perfect Rectangle

问题：

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.

解决：

【题意】这道题是说给了一堆小矩形的坐标（左下角和右上角围成的），问其可否组合成一个完美的矩形（刚恰好，很少，很多，不交叉重复）。

① 核心思想：可以正好围成一个矩形的状况就是:

 有且只有：

- 最左下 最左上 最右下 最右上 的四个点只出现过一次,其余确定是成对出现的(保证彻底覆盖)

- 上面四个点围成的面积,正好等于全部子矩形的面积之和(保证不重复)

public class Solution { //106ms
    public boolean isRectangleCover(int[][] rectangles) {
        int x1 = Integer.MAX_VALUE;
        int x2 = Integer.MIN_VALUE;
        int y1 = Integer.MAX_VALUE;
        int y2 = Integer.MIN_VALUE;
        int area = 0;
        Set<String> set = new HashSet();
        for(int[] rect : rectangles) {
            area += (rect[2] - rect[0]) * (rect[3] - rect[1]);
            x1 = Math.min(x1, rect[0]);
            y1 = Math.min(y1, rect[1]);
            x2 = Math.max(x2, rect[2]);
            y2 = Math.max(y2, rect[3]);
            String s1 = rect[0] + " " + rect[1];
            String s2 = rect[0] + " " + rect[3];
            String s3 = rect[2] + " " + rect[1];
            String s4 = rect[2] + " " + rect[3];
            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);
        }
        // condition 1
        if(area != (x2 - x1) * (y2 - y1)) return false;
        // condition 2
        return set.contains(x1 + " " + y1) && set.contains(x1 + " " + y2) && set.contains(x2 + " " + y1) && set.contains(x2 + " " + y2) && set.size() == 4;

    } }