# 478. Generate Random Point in a Circle (Medium)

https://leetcode.com/problems/generate-random-point-in-a-circle/

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.

</div>

## Solutions

``````class Solution {

double r;
double x;
double y;

public Solution(double radius, double x_center, double y_center) {
x = x_center;
y = y_center;
}

public double[] randPoint() {
double rx = 0;
double ry = 0;
int maxInt = (1 << 31) - 1;

Random rand = new Random();
do {
// The special case of below statements is that the circle center is never covered.
// The reason is that Math.random returns a value range in [0, 1), 0 inclusive 1 exclusive.

// rx = x + (1 - Math.random()) * r;
// ry = y + (1 - Math.random()) * r'

// The nextInt() method of Random accepts a bound integer and returns a random integer
// from 0 (inclusive) to the specified bound (exclusive). By default, the specified bound
// is (1<<31)-1

// To make rand.nextInt() produce values in [0,1], both 0 and 1 are inclusive, use
// following lines. Don't forget to deduct 1 from maxInt since the maximum rand.nextInt()
// can reach is maxInt-1.
rx = x + rand.nextInt() * 1.0 / (maxInt - 1) * r;
ry = y + rand.nextInt() * 1.0 / (maxInt - 1) * r;

} while (Math.pow(rx - x, 2) + Math.pow(ry - y, 2) > Math.pow(r, 2));

// Points out of the boundary of the circle are rejected.

return new double[]{rx, ry};
}
}
``````