Random Point in Non-overlapping Rectangles (M)

題目

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.

題意

給定若干個互不重疊的矩形，要求每次從這些矩形覆蓋的區域中均勻隨機地選出一個點。

思路

我們不能單純的給每個矩形編號，隨機選一個編號再從對應的矩形中隨機選一個點，因為這樣不能保證每個點被選到的概率是一樣的。應該以每個矩形覆蓋的面積來確定它被選到的概率。具體操作為：遍歷所有矩形，每次累加當前矩形包含的點數(即面積)，並以此作為當前矩形的編號，可以得到一個類似數軸的圖：

記總點數為area，從區間[1, area]中隨機選一個點p，那麼對應的矩形編號就是p所在的顏色方塊的右端點。再在這個矩形中隨機選出一個點。以這種方式選取矩形能夠保證每個點被選到的概率是一樣的。

程式碼實現

Java

class Solution {
    private Random random;
    // 為了方便使用了TreeMap，也可以用HashMap結合二分搜尋找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();
 */