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)).

Tips：

O(n)

两条规则： 1.大的长方形面积必须等于所有小长方形面积之和2. 2.正方形四个角必须只出现过一次

Code：

public class Solution {
    public boolean isRectangleCover(int[][] rectangles) {

        if (rectangles.length == 0 || rectangles[0].length == 0) return false;

        int x1 = Integer.MAX_VALUE;
        int x2 = Integer.MIN_VALUE;
        int y1 = Integer.MAX_VALUE;
        int y2 = Integer.MIN_VALUE;

        HashSet<String> set = new HashSet<String>();
        int area = 0;

        for (int[] rect : rectangles) {
            x1 = Math.min(rect[0], x1);
            y1 = Math.min(rect[1], y1);
            x2 = Math.max(rect[2], x2);
            y2 = Math.max(rect[3], y2);

            area += (rect[2] - rect[0]) * (rect[3] - rect[1]);

            String s1 = rect[0] + " " + rect[1];
            String s2 = rect[0] + " " + rect[3];
            String s3 = rect[2] + " " + rect[3];
            String s4 = rect[2] + " " + rect[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(x1 + " " + y1) || !set.contains(x1 + " " + y2) || !set.contains(x2 + " " + y1) || !set.contains(x2 + " " + y2) || set.size() != 4) return false;

        return area == (x2-x1) * (y2-y1);
    }
}