LeetcodeOJ 391 ：Perfect Rectangle

题目地址：https://leetcode.com/problems/perfect-rectangle/description/

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.

开始的想法是建立一个board格子，记录哪些格子会被矩形覆盖且只覆盖一次，同时记录生成大矩形的左下角和右上角的坐标，计算小矩形面积之和与最后的大矩形面积是否相等。该方法在数据较大的时候会超时.

代码如下：

class Solution {    public boolean isRectangleCover(int[][] rectangles) {int down = Integer.MAX_VALUE;int left = Integer.MAX_VALUE;int up = Integer.MIN_VALUE;int right = Integer.MIN_VALUE;for(int[] item : rectangles) {left = Math.min(left, item[0]);down = Math.min(down, item[1]);right = Math.max(right, item[2]);up = Math.max(up, item[3]);}int MAX = Math.max(up, right);int MIN = Math.min(down, left);int[][] board = new int[MAX+1-MIN][MAX+1-MIN];for(int[] item : rectangles) {for(int i = item[0]-MIN; i < item[2]-MIN; ++i) {for(int j = item[1]-MIN; j < item[3]-MIN; ++j) {if(board[i][j] == 1)return false;board[i][j] = 1;}}}for(int i = left-MIN; i < right-MIN; ++i) {for(int j = down-MIN; j < up-MIN; ++j) {if(board[i][j] != 1)return false;}}return true;    }}

改进方法核心思想是:能够正好围成一个矩形的情况就是: 

1. 最左下 最左上 最右下 最右上 的四个点只出现过一次,其他肯定是成对出现的(保证完全覆盖) 
2. 上面四个点围成的面积,正好等于所有子矩形的面积之和(保证不重复)

代码如下：

class Solution {    public boolean isRectangleCover(int[][] rectangles) {        int down = Integer.MAX_VALUE;int left = Integer.MAX_VALUE;int up = Integer.MIN_VALUE;int right = Integer.MIN_VALUE;int sum = 0; // 面积Set<String> set = new HashSet<>();for(int[] item : rectangles) {//四个顶点坐标String s1 = item[0] + " " + item[1];String s2 = item[2] + " " + item[3];String s3 = item[0] + " " + item[3];String s4 = item[2] + " " + item[1];if(set.contains(s1)) set.remove(s1);elseset.add(s1);if(set.contains(s2)) set.remove(s2);elseset.add(s2);if(set.contains(s3)) set.remove(s3);elseset.add(s3);if(set.contains(s4)) set.remove(s4);elseset.add(s4);down = Math.min(down, item[1]);left = Math.min(left, item[0]);up = Math.max(up, item[3]);right = Math.max(right, item[2]);sum += (item[3]-item[1]) * (item[2]-item[0]);}int area = (up-down) * (right-left);if(area != sum || set.size() != 4 || !set.contains(left+" "+down) || !set.contains(left+" "+up) || !set.contains(right+" "+down) || !set.contains(right+" "+up))return false;return true;    }}