0478. Generate Random Point in a Circle (M)

Generate Random Point in a Circle (M)

题目

Given the radius and x-y positions of the center of a circle, write a function randPoint which generates a uniform random point in the circle.

Note:

1. input and output values are in floating-point.
2. radius and x-y position of the center of the circle is passed into the class constructor.
3. a point on the circumference of the circle is considered to be in the circle.
4. randPoint returns a size 2 array containing x-position and y-position of the random point, in that order.

Example 1:

Input: 
["Solution","randPoint","randPoint","randPoint"]
[[1,0,0],[],[],[]]
Output: [null,[-0.72939,-0.65505],[-0.78502,-0.28626],[-0.83119,-0.19803]]

Example 2:

Input: 
["Solution","randPoint","randPoint","randPoint"]
[[10,5,-7.5],[],[],[]]
Output: [null,[11.52438,-8.33273],[2.46992,-16.21705],[11.13430,-12.42337]]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution's constructor has three arguments, the radius, x-position of the center, and y-position of the center of the circle. randPoint has no arguments. Arguments are always wrapped with a list, even if there aren't any.

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题意

在圆内随机生成一个点。

思路

极坐标容易处理：生成一个随机的角度rad，一个随机的长度len，len在0-radius之间，那么得到的随机点就是(len*cos(rad) + x, len*sin(rad) + y)。问题在于如果len直接由radius乘一个0-1的随机数random生成，最后得到的结果并不是对于面积等概率的。举个例子，如果要在半径为R的圆中取一点，使该点正好在半径为R/2的同心圆中，概率为1/4而不是1/2。因此应当先对random去平方根，再与半径相乘来得到len值。

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代码实现

Java

class Solution {
    private double r, x, y;
    private Random random;

    public Solution(double radius, double x_center, double y_center) {
        r = radius;
        x = x_center;
        y = y_center;
        random = new Random();
    }

    public double[] randPoint() {
        double len = Math.sqrt(random.nextDouble()) * r;
        double rad = random.nextDouble() * Math.PI * 2;
        return new double[]{len * Math.cos(rad) + x, len * Math.sin(rad) + y};
    }
}