leetcode 850. Rectangle Area II-优快云博客

该博客讨论了LeetCode第850题，题目要求计算一组轴对齐矩形在平面上覆盖的总面积，并给出了解题思路，包括一种巧妙的解法和线段树的解决方案。

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We are given a list of (axis-aligned) rectangles. Each rectangle[i] = [x1, y1, x2, y2] , where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the ith rectangle.

Find the total area covered by all rectangles in the plane. Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
Output: 6
Explanation: As illustrated in the picture.

Example 2:

Input: [[0,0,1000000000,1000000000]]
Output: 49
Explanation: The answer is 10^18 modulo (10^9 + 7), which is (10^9)^2 = (-7)^2 = 49.

Note:

1 <= rectangles.length <= 200
rectanges[i].length = 4
0 <= rectangles[i][j] <= 10^9
The total area covered by all rectangles will never exceed 2^63 - 1 and thus will fit in a 64-bit signed integer.

解题思路：

没做出来。。参考leetcode大神的解法，真的很巧妙，对求矩形面积很实用；

class Solution {
public:
    int rectangleArea(vector<vector<int>>& rectangles) 
    {
        set<int> ys ;
        long res = 0 , modl = 1000000007 ;
        
        for(auto& rec : rectangles)
        {
            ys.insert(rec[1]) ;
            ys.insert(rec[3]) ;
        }
        
        sort(rectangles.begin() , rectangles.end() , [](auto &a , auto &b){return a[0] < b[0] ;}) ;
        
        int prev_y = *ys.begin() ;
        
        for(auto y : ys)
        {
            long height = y - prev_y ;
            long x_start = rectangles.front()[0] ;
            long x_end = x_start ;
            
            for(auto &rec : rectangles)
            {
                if(rec[1] <= prev_y && rec[3] >= y )
                {
                    if(rec[0] > x_end)
                    {
                        res = (res + (x_end - x_start) * height ) % modl ;
                        x_start = rec[0] ;
                    }
                    if(rec[2] > x_end)
                    {
                        x_end = rec[2] ;
                    }
                }
            }
            
            res = (res + (x_end - x_start) * height ) % modl ;
            prev_y = y ;
        }
        
        return res ;
    }
};

解法二：

还有一种线段树的做法。。。