478 Generate Random Point in a Circle

1. Question

Given the radius and x-y positions of the center of a circle, write a functionrandPoint 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.randPointhas no arguments. Arguments are always wrapped with a list, even if there aren't any.

2. Implementation

(1) 用极坐标表示

思路:

class Solution {
    double radius, x_center, y_center;

    public Solution(double radius, double x_center, double y_center) {
        this.radius = radius;
        this.x_center = x_center;
        this.y_center = y_center;
    }

    public double[] randPoint() {
        double len = Math.sqrt(Math.random()) * radius;
        int degree = (int)(Math.random() * 360);
        double x = x_center + len * Math.cos(degree);
        double y = y_center + len * Math.sin(degree);
        return new double[] {x, y};
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(radius, x_center, y_center);
 * double[] param_1 = obj.randPoint();
 */

3. Time & Space Complexity

时间复杂度O(1), 空间复杂度O(1)

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