478. Generate Random Point in a Circle

Problem

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.

Problem Summary

Given the radius and the x and y coordinates of the center of a circle, write a function randPoint that generates a uniformly random point inside the circle.

Notes:

• Input values and output values will all be floating-point numbers.
• The radius of the circle and the x and y coordinates of the center will be passed as parameters to the class constructor.
• Points on the circumference are also considered to be inside the circle.
• randPoint returns an array of size 2 containing the x-coordinate and y-coordinate of the random point.

Solution Approach

• Randomly generate a point inside the circle. This point must satisfy the definition (x-a)^2+(y-b)^2 ≤ R^2, where (a,b) is the coordinate of the center of the circle, and R is the radius.
• First assume the center of the circle is at (0,0), which makes calculation easier. When finally outputting the coordinates, simply add the offset of the center as a whole. rand.Float64() generates a floating-point number in the range [0.0,1.0). -R ≤ 2 * R * rand() - R < R. Use the relationship between the randomly generated point's horizontal and vertical coordinates (x,y) and the radius R. If x^2 + y^2 ≤ R^2, then the generated point is inside the circle. When finally outputting, remember to add the offset value of the center coordinates.

Code

package leetcode

import (
	"math"
	"math/rand"
	"time"
)

type Solution struct {
	r float64
	x float64
	y float64
}

func Constructor(radius float64, x_center float64, y_center float64) Solution {
	rand.Seed(time.Now().UnixNano())
	return Solution{radius, x_center, y_center}
}

func (this *Solution) RandPoint() []float64 {
	/*
	   a := angle()
	   r := this.r * math.Sqrt(rand.Float64())
	   x := r * math.Cos(a) + this.x
	   y := r * math.Sin(a) + this.y
	   return []float64{x, y}*/
	for {
		rx := 2*rand.Float64() - 1.0
		ry := 2*rand.Float64() - 1.0
		x := this.r * rx
		y := this.r * ry
		if x*x+y*y <= this.r*this.r {
			return []float64{x + this.x, y + this.y}
		}
	}
}

func angle() float64 {
	return rand.Float64() * 2 * math.Pi
}

/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(radius, x_center, y_center);
 * param_1 := obj.RandPoint();
 */