497. Random Point in Non-overlapping Rectangles

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.

https://leetcode.com/problems/random-point-in-non-overlapping-rectangles/discuss/316890/Trying-to-explain-why-the-intuitive-solution-wont-work

以开始想到随机挑选一个rec然后随机选里面的点不就行了？但这样是不行的，因为rec面积不同，但你选他们的概率却是相同的，这样不符合random，因为对大面积的rec不公平。上面的链接解释了这一情况

class Solution {
    TreeMap<Integer, Integer> map;
    int[][] arrays;
    int sum;
    Random rnd= new Random();
    
    public Solution(int[][] rects) {
        arrays = rects;
        map = new TreeMap<>();
        sum = 0;
        
        for(int i = 0; i < rects.length; i++) {
            int[] rect = rects[i];
                        
            // the right part means the number of points can be picked in this rectangle
            sum += (rect[2] - rect[0] + 1) * (rect[3] - rect[1] + 1);
            
            map.put(sum, i);
        }
    }
    
    public int[] pick() {
        // nextInt(sum) returns a num in [0, sum -1]. After added by 1, it becomes [1, sum]
        int c = map.ceilingKey( rnd.nextInt(sum) + 1);
        
        return pickInRect(arrays[map.get(c)]);
    }
    
    private int[] pickInRect(int[] rect) {
        int left = rect[0], right = rect[2], bot = rect[1], top = rect[3];
        
        return new int[]{left + rnd.nextInt(right - left + 1), bot + rnd.nextInt(top - bot + 1) };
    }
}

正确的方法：

像presum一样，把presum和对应的数组index存到map中，然后rand.nextInt(sum) + 1代表当前取到的随机sum，然后看他在哪个recindex里，随即返回那个rec里的一个点即可。

ceilingKey：return a key larger or equal than parameter.