391. Perfect Rectangle

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LeetCode

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Problem

Given an array rectangles where rectangles[i] = [xi, yi, ai, bi] represents an axis-aligned rectangle. The bottom-left point of the rectangle is (xi, yi) and the top-right point of it is (ai, bi).

Return true if all the rectangles together form an exact cover of a rectangular region.

Example 1:

Input: rectangles = [[1,1,3,3],[3,1,4,2],[3,2,4,4],[1,3,2,4],[2,3,3,4]]
Output: true
Explanation: All 5 rectangles together form an exact cover of a rectangular region.

Example 2:

Input: rectangles = [[1,1,2,3],[1,3,2,4],[3,1,4,2],[3,2,4,4]]
Output: false
Explanation: Because there is a gap between the two rectangular regions.

Example 3:

Input: rectangles = [[1,1,3,3],[3,1,4,2],[1,3,2,4],[2,2,4,4]]
Output: false
Explanation: Because two of the rectangles overlap with each other.

Code

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;
        
        // 放入hashset，第一次遇见加入，第二次遇见删除
        // 最后剩下顶点
        HashSet<String> set = new HashSet<>();
        
        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);
    }
}