[LeetCode] Random Point in Non-overlapping Rectangles 非重叠矩形中的随机点

 

Given a list of non-overlapping axis-aligned rectangles rects, write a function pick which randomly and uniformily picks an integer point in the space covered by the rectangles.

Note:

1. An integer point is a point that has integer coordinates. 
2. A point on the perimeter of a rectangle is included in the space covered by the rectangles. 
3. ith rectangle = rects[i] = [x1,y1,x2,y2], where [x1, y1] are the integer coordinates of the bottom-left corner, and [x2, y2] are the integer coordinates of the top-right corner.
4. length and width of each rectangle does not exceed 2000.
5. 1 <= rects.length <= 100
6. pick return a point as an array of integer coordinates [p_x, p_y]
7. pick is called at most 10000 times.

Example 1:

Input: 
["Solution","pick","pick","pick"]
[[[[1,1,5,5]]],[],[],[]] Output: [null,[4,1],[4,1],[3,3]] 

Example 2:

Input: 
["Solution","pick","pick","pick","pick","pick"]
[[[[-2,-2,-1,-1],[1,0,3,0]]],[],[],[],[],[]] Output: [null,[-1,-2],[2,0],[-2,-1],[3,0],[-2,-2]]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution's constructor has one argument, the array of rectangles rects. pick has no arguments. Arguments are always wrapped with a list, even if there aren't any.

 

这道题给了咱们一些非重叠的矩形，让咱们返回一个这些矩形中的一个随机的点。那么博主的第一直觉就是首先要在这些矩形中随机挑出来一个，而后在这个随机的矩形中再随机生成一个点，经过随机生成一个长和宽便可。博主最开始想到的方法是用rand随机生成一个 [0, n) 范围内的数字，n为输入矩形的个数，这样就获得了一个随机的矩形。可是这种方法貌似行不通，会跪在一个很长的输入测试数据。这使得博主比较困惑了，没有想出缘由是为什么，有哪位看官大神们知道的，麻烦留言告知博主哈！哈，已经知道了，参见评论区二楼留言～ 论坛上的解法有一种是用水塘抽样Reservoir Sampling的方法的，LeetCode以前有过几道须要用这种方法的题目 Random Pick Index，Shuffle an Array 和 Linked List Random Node。这里咱们使用其来随机出一个矩形，作法是遍历全部的矩形，用变量sumArea来计算当前遍历过的全部矩形面积之和，而后变量area是当前遍历的矩形的面积（注意这里不是严格意义上的面积，而是该区域内整数坐标的点的个数，因此长宽相乘的时候都要加1），而后咱们在当前全部矩形面积之和内随机生成一个值，若是这个值小于area，那么选择当前的矩阵为随机矩形。这里至关于一个大小为area的水塘，在这个值以内的话，就更换selected。这个方法是没啥问题，可是博主仍是没想通为啥不能直接随机生成矩形的index。当咱们拿到随机矩形后，以后就随机出宽和高返回便可，参见代码以下：

 

解法一：

class Solution {
public:
    Solution(vector<vector<int>> rects) {
        _rects = rects;
    }
    
    vector<int> pick() {
        int sumArea = 0;
        vector<int> selected;
        for (auto rect : _rects) {
            int area = (rect[2] - rect[0] + 1) * (rect[3] - rect[1] + 1);
            sumArea += area;
            if (rand() % sumArea < area) selected = rect;
        }
        int x = rand() % (selected[2] - selected[0] + 1) + selected[0];
        int y = rand() % (selected[3] - selected[1] + 1) + selected[1];
        return {x, y};
    }

private:
    vector<vector<int>> _rects;
};

 

这道题在论坛上的主流解法实际上是这个，咱们用TreeMap来创建当前遍历过的矩形面积之和跟该矩形位置之间的映射。而后当咱们求出全部的矩形面积之和后，咱们随机生成一个值，而后在TreeMap中找到第一个大于这个值的矩形，这里博主仍是有疑问，为啥不能直接随机矩形的位置，而是非要跟面积扯上关系。以后的步骤就跟上面的没啥区别了，参见代码以下：

 

解法二：

class Solution {
public:
    Solution(vector<vector<int>> rects) {
        _rects = rects;
        _totalArea = 0;
        for (auto rect : rects) {
            _totalArea += (rect[2] - rect[0] + 1) * (rect[3] - rect[1] + 1);
            _areaToIdx.insert({_totalArea, _areaToIdx.size()});
        }
    }
    
    vector<int> pick() {
        int val = rand() % _totalArea;
        int idx = _areaToIdx.upper_bound(val)->second;
        int width = _rects[idx][2] - _rects[idx][0] + 1;
        int height = _rects[idx][3] - _rects[idx][1] + 1;
        return {rand() % width + _rects[idx][0], rand() % height + _rects[idx][1]};
    }

private:
    vector<vector<int>> _rects;
    int _totalArea;
    map<int, int> _areaToIdx;
};

 

相似题目：

Random Pick with Weight

Generate Random Point in a Circle

 

参考资料：

https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/

https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/discuss/155005/C%2B%2B-single-rand()-call

https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/discuss/169185/Short-C%2B%2B-solution-with-upper_bound

https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/discuss/170503/C%2B%2B-solution-using-reservoir-sampling-with-explanation-concise-and-easy-to-understand

 

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