0497. Random Point in Non-overlapping Rectangles (M)

Random Point in Non-overlapping Rectangles (M)

题目

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.

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题意

给定若干个互不重叠的矩形，要求每次从这些矩形覆盖的区域中均匀随机地选出一个点。

思路

我们不能单纯的给每个矩形编号，随机选一个编号再从对应的矩形中随机选一个点，因为这样不能保证每个点被选到的概率是一样的。应该以每个矩形覆盖的面积来确定它被选到的概率。具体操作为：遍历所有矩形，每次累加当前矩形包含的点数(即面积)，并以此作为当前矩形的编号，可以得到一个类似数轴的图：

记总点数为area，从区间[1, area]中随机选一个点p，那么对应的矩形编号就是p所在的颜色方块的右端点。再在这个矩形中随机选出一个点。以这种方式选取矩形能够保证每个点被选到的概率是一样的。

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代码实现

Java

class Solution {
    private Random random;
    // 为了方便使用了TreeMap，也可以用HashMap结合二分搜索找key
    private TreeMap<Integer, int[]> map;
    private int area;

    public Solution(int[][] rects) {
        random = new Random();
        map = new TreeMap<>();
        for (int[] rect : rects) {
            area += (rect[2] - rect[0] + 1) * (rect[3] - rect[1] + 1);
            map.put(area, rect);
        }
    }

    public int[] pick() {
        int[] rect = map.get(map.ceilingKey(random.nextInt(area) + 1));
        int x = rect[0] + random.nextInt(rect[2] - rect[0] + 1);
        int y = rect[1] + random.nextInt(rect[3] - rect[1] + 1);
        return new int[] { x, y };
    }
}

/**
 * Your Solution object will be instantiated and called as such: Solution obj =
 * new Solution(rects); int[] param_1 = obj.pick();
 */