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:
- An integer point is a point that has integer coordinates.
- A point on the perimeter of a rectangle is included in the space covered by the rectangles.
-
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. - length and width of each rectangle does not exceed
2000. 1 <= rects.length <= 100-
pickreturn a point as an array of integer coordinates[p_x, p_y] -
pickis called at most10000times.
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.
这道题给了我们一些非重叠的矩形，让我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，然后在这个随机的矩形中再随机生成一个点，通过随机生成一个长和宽即可。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，这样就得到了一个随机的矩形。但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，没有想出原因是为何，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之前有过几道需要用这种方法的题目 Random Pick Index，Shuffle an Array 和 Linked List Random Node。这里我们使用其来随机出一个矩形，做法是遍历所有的矩形，用变量sumArea来计算当前遍历过的所有矩形面积之和，然后变量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，所以长宽相乘的时候都要加1），然后我们在当前所有矩形面积之和内随机生成一个值，如果这个值小于area，那么选择当前的矩阵为随机矩形。这里相当于一个大小为area的水塘，在这个值之内的话，就更换selected。这个方法是没啥问题，但是博主还是没想通为啥不能直接随机生成矩形的index。当我们拿到随机矩形后，之后就随机出宽和高返回即可，参见代码如下：
解法一：
class Solution {
public:
Solution(vector<vector<int>> rects) {
_rects = rects;
}
vector<int> pick() {
int sumArea = ;
vector<int> selected;
for (auto rect : _rects) {
int area = (rect[] - rect[] + ) * (rect[] - rect[] + );
sumArea += area;
if (rand() % sumArea < area) selected = rect;
}
int x = rand() % (selected[] - selected[] + ) + selected[];
int y = rand() % (selected[] - selected[] + ) + selected[];
return {x, y};
}
private:
vector<vector<int>> _rects;
};
这道题在论坛上的主流解法其实是这个，我们用TreeMap来建立当前遍历过的矩形面积之和跟该矩形位置之间的映射。然后当我们求出所有的矩形面积之和后，我们随机生成一个值，然后在TreeMap中找到第一个大于这个值的矩形，这里博主还是有疑问，为啥不能直接随机矩形的位置，而是非要跟面积扯上关系。之后的步骤就跟上面的没啥区别了，参见代码如下：
解法二：
class Solution {
public:
Solution(vector<vector<int>> rects) {
_rects = rects;
_totalArea = ;
for (auto rect : rects) {
_totalArea += (rect[] - rect[] + ) * (rect[] - rect[] + );
_areaToIdx.insert({_totalArea, _areaToIdx.size()});
}
}
vector<int> pick() {
int val = rand() % _totalArea;
int idx = _areaToIdx.upper_bound(val)->second;
int width = _rects[idx][] - _rects[idx][] + ;
int height = _rects[idx][] - _rects[idx][] + ;
return {rand() % width + _rects[idx][], rand() % height + _rects[idx][]};
}
private:
vector<vector<int>> _rects;
int _totalArea;
map<int, int> _areaToIdx;
};
类似题目：
Generate Random Point in a Circle
参考资料：
https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/