Skip to content

Latest commit

 

History

History
140 lines (110 loc) · 3.01 KB

File metadata and controls

140 lines (110 loc) · 3.01 KB

中文文档

Description

There is a special square room with mirrors on each of the four walls. Except for the southwest corner, there are receptors on each of the remaining corners, numbered 0, 1, and 2.

The square room has walls of length p and a laser ray from the southwest corner first meets the east wall at a distance q from the 0th receptor.

Given the two integers p and q, return the number of the receptor that the ray meets first.

The test cases are guaranteed so that the ray will meet a receptor eventually.

 

Example 1:

Input: p = 2, q = 1
Output: 2
Explanation: The ray meets receptor 2 the first time it gets reflected back to the left wall.

Example 2:

Input: p = 3, q = 1
Output: 1

 

Constraints:

  • 1 <= q <= p <= 1000

Solutions

Python3

class Solution:
    def mirrorReflection(self, p: int, q: int) -> int:
        g = gcd(p, q)
        p = (p // g) % 2
        q = (q // g) % 2
        if p == 1 and q == 1:
            return 1
        return 0 if p == 1 else 2

Java

class Solution {
    public int mirrorReflection(int p, int q) {
        int g = gcd(p, q);
        p = (p / g) % 2;
        q = (q / g) % 2;
        if (p == 1 && q == 1) {
            return 1;
        }
        return p == 1 ? 0 : 2;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

C++

class Solution {
public:
    int mirrorReflection(int p, int q) {
        int g = __gcd(p, q);
        p = (p / g) % 2;
        q = (q / g) % 2;
        if (p == 1 && q == 1) {
            return 1;
        }
        return p == 1 ? 0 : 2;
    }
};

Go

func mirrorReflection(p int, q int) int {
	g := gcd(p, q)
	p = (p / g) % 2
	q = (q / g) % 2
	if p == 1 && q == 1 {
		return 1
	}
	if p == 1 {
		return 0
	}
	return 2
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

TypeScript

function mirrorReflection(p: number, q: number): number {
    const g = gcd(p, q);
    p = Math.floor(p / g) % 2;
    q = Math.floor(q / g) % 2;
    if (p === 1 && q === 1) {
        return 1;
    }
    return p === 1 ? 0 : 2;
}

function gcd(a: number, b: number): number {
    return b === 0 ? a : gcd(b, a % b);
}

...