[leetcode] 478. Generate Random Point in a Circle

[leetcode] 478. Generate Random Point in a Circle

Description

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:

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.

分析

题目的意思是：这道题给定一个圆，随机返回圆内的一个点。我最开始用random.random测了一下，发现并没有完全ac。看来问题并不简单。我参考了一下别人的思路，其中x1的random区间为[-1,1],y1的random区间为[-1,1]，那最终的坐标就可以算为：

[self.x_center+x1*self.radius,self.y_center+y1*self.radius]

x1和y1乘以半径，然后加上中心坐标就行了。

代码

class Solution: def __init__(self, radius: float, x_center: float, y_center: float): self.radius=radius self.x_center=x_center self.y_center=y_center def randPoint(self) -> List[float]: while(True): x1=(random.random()-0.5)*2 y1=(random.random()-0.5)*2 if(x1**2+y1**2<=1): return [self.x_center+x1*self.radius,self.y_center+y1*self.radius] # Your Solution object will be instantiated and called as such:# obj = Solution(radius, x_center, y_center)# param_1 = obj.randPoint()

参考文献

​​Simple Python solution with minimum library use (92% Faster)​​

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。