497 Random Point in Non-overlapping Rectangles 非重叠矩形中的随机点
Description:
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
pick return a point as an array of integer coordinates [p_x, p_y]
pick is called at most 10000 times.
Example:
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.
题目描述:
给定一个非重叠轴对齐矩形的列表 rects，写一个函数 pick 随机均匀地选取矩形覆盖的空间中的整数点。
提示:
整数点是具有整数坐标的点。
矩形周边上的点包含在矩形覆盖的空间中。
第 i 个矩形 rects [i] = [x1，y1，x2，y2]，其中 [x1，y1] 是左下角的整数坐标，[x2，y2] 是右上角的整数坐标。
每个矩形的长度和宽度不超过 2000。
1 <= rects.length <= 100
pick 以整数坐标数组 [p_x, p_y] 的形式返回一个点。
pick 最多被调用10000次。
示例 :
示例 1：
输入:
["Solution","pick","pick","pick"]
[[[[1,1,5,5]]],[],[],[]]
输出:
[null,[4,1],[4,1],[3,3]]
示例 2：
输入:
["Solution","pick","pick","pick","pick","pick"]
[[[[-2,-2,-1,-1],[1,0,3,0]]],[],[],[],[],[]]
输出:
[null,[-1,-2],[2,0],[-2,-1],[3,0],[-2,-2]]
输入语法的说明：
输入是两个列表：调用的子例程及其参数。Solution 的构造函数有一个参数，即矩形数组 rects。pick 没有参数。参数总是用列表包装的，即使没有也是如此。
思路:
求每个矩阵的点数及点数的前缀和
然后用二分法求出矩阵的下标即可
预处理时间复杂度O(n), pick()时间复杂度O(lgn), 空间复杂度O(1)
代码:
C++:
class Solution
{
private:
int area;
map<int, int> m;
vector<vector<int>> rects;
public:
Solution(vector<vector<int>>& rects)
{
this -> rects = rects;
int sum = 0;
for (int i = 0; i < rects.size(); i++)
{
int cur = (rects[i][2] - rects[i][0] + 1) * (rects[i][3] - rects[i][1] + 1);
sum += cur;
m[sum] = i;
}
area = sum;
}
vector<int> pick()
{
random_device rd;
default_random_engine e(rd());
uniform_int_distribution dis(0, area);
int index = m.upper_bound(dis(e)) -> second;
uniform_int_distribution x(rects[index][0], rects[index][2]), y(rects[index][1], rects[index][3]);
return { x(e), y(e) };
}
};
/**
* Your Solution object will be instantiated and called as such:
* Solution* obj = new Solution(rects);
* vector<int> param_1 = obj->pick();
*/
Java:
class Solution {
private int[][] rects;
private int len;
private int pre[];
private Random r = new Random();
public Solution(int[][] rects) {
pre = new int[rects.length + 1];
for (int i = 0; i < rects.length; i++) pre[i + 1] += pre[i] + (rects[i][2]-rects[i][0] + 1) * (rects[i][3] - rects[i][1] + 1);
this.rects = rects;
len = pre[pre.length - 1];
}
public int[] pick() {
int p = r.nextInt(len), left = 1, right = pre.length - 1;
while (left < right) {
int mid = left + ((right - left) >>> 1);
if (p >= pre[mid]) left = mid + 1;
else right = mid;
}
int index = p - pre[left - 1], col = rects[left - 1][3] - rects[left - 1][1] + 1;
return new int[] { index / col + rects[left - 1][0], index % col + rects[left - 1][1] };
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(rects);
* int[] param_1 = obj.pick();
*/
Python:
class Solution:
def __init__(self, rects: List[List[int]]):
self.rects = rects
self.weight = []
s = 0
for rect in rects:
area = (rect[2] - rect[0] + 1) * (rect[3] - rect[1] + 1)
s += area
self.weight.append(s)
def pick(self) -> List[int]:
index = bisect.bisect_left(self.weight, randint(1, self.weight[-1]))
rect = self.rects[index]
return [random.randint(rect[0], rect[2]), random.randint(rect[1], rect[3])]
# Your Solution object will be instantiated and called as such:
# obj = Solution(rects)
# param_1 = obj.pick()
