[Leetcode] 391. Perfect Rectangle 解题报告

最新推荐文章于 2021-11-18 16:46:21 发布

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本文介绍了一种高效算法来判断多个轴对齐的矩形是否能完全覆盖一个更大的矩形区域，且没有重叠或空缺。通过使用哈希表记录矩形顶点的状态，该算法能在O(n)时间内解决问题。

题目：

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.

思路：

自己的笨办法就是先计算所有小矩阵的面积和，以及四个边界构成的面积，如果不相等就直接返回false。接下来就需要判断是否有重合：如果要求空间复杂度是O(1)，就可以两两检查矩形，这样时间复杂度是O(n^2)；如果允许空间复杂度是O(n^2)，那么可以初始化四个边界构成的矩阵，然后把每个矩阵“往上贴”，贴的过程中如果发现某个位置已经被其他矩阵覆盖，则返回false。当然这种算法不能通过大数据测试，我们需要更聪明的方法。

小榕流光的这篇帖子http://blog.youkuaiyun.com/qq508618087/article/details/52483625图文并茂，解释的很详细，为了不构成侵权，请参考我给出的网址^_^。这种算法的时间复杂度是O(n)。

代码：

class Solution {
public:
    bool isRectangleCover(vector<vector<int>>& rectangles) {
        unordered_map<string, int> hash;  
        for(auto val: rectangles) {  
            for(int i = 0; i < 4; i++) {  
                string tem = to_string(val[i / 2 * 2]) + ','+to_string(val[i % 2 * 2 + 1]);  
                if(hash[tem] & (1 << i)) {            // already another corner with the same index
                    return false;  
                }
                hash[tem] |= (1 << i);  
            }  
        }  
        int cntCorner = 0;  
        for(auto& val: hash) {  
            int sec = val.second;  
            if(!(sec & (sec - 1)) && cntCorner++ > 4)  {                        // When the point is the corner
                return false;
            }
            if((sec&(sec-1)) && !(sec==3||sec==12||sec==5||sec==10||sec==15)) { // when the point is not the corner
                return false; 
            }
        }  
        return true;  
    }
};