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:
input and output values are in floating-point.
radius and x-y position of the center of the circle is passed into the class constructor.
a point on the circumference of the circle is considered to be in the circle.
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.
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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