[Swift]LeetCode365. 水壶问题 | Water and Jug Problem

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You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

• Fill any of the jugs completely with water.
• Empty any of the jugs.
• Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

Example 1: (From the famous "Die Hard" example)

Input: x = 3, y = 5, z = 4
Output: True

Example 2:

Input: x = 2, y = 6, z = 5
Output: False

---

有两个容量分别为 x升 和 y升 的水壶以及无限多的水。请判断能否通过使用这两个水壶，从而可以得到恰好 z升 的水？

如果可以，最后请用以上水壶中的一或两个来盛放取得的 z升 水。

你允许：

• 装满任意一个水壶
• 清空任意一个水壶
• 从一个水壶向另外一个水壶倒水，直到装满或者倒空

示例 1: (From the famous "Die Hard" example)

输入: x = 3, y = 5, z = 4
输出: True

示例 2:

输入: x = 2, y = 6, z = 5
输出: False

---

8ms

class Solution {
    func canMeasureWater(_ x: Int, _ y: Int, _ z: Int) -> Bool {
        return z == 0 || (x + y >= z && z % gcd(x, y) == 0)
    }
    
    func gcd(_ x:Int,_ y:Int) -> Int
    {
        return y == 0 ? x : gcd(y, x % y)
    }
}

---

8ms

class Solution {
    func canMeasureWater(_ x: Int, _ y: Int, _ z: Int) -> Bool {
        guard z <= x + y else {
            return false
        }
        
        let divisor = greatestCommonDivisor(x, y)
        
        return z == 0 ? true : z % divisor == 0
    }
    
    func greatestCommonDivisor(_ x: Int, _ y: Int) -> Int {
        
        if y == 0 {
            return x
        }
        
        return greatestCommonDivisor(y, x % y)
    }
}

---

12ms

class Solution {
    func canMeasureWater(_ x: Int, _ y: Int, _ z: Int) -> Bool {
        func gcd(x: Int, y: Int) -> Int {
            if y == 0 {
                return x
            } else {
                return gcd(x: y, y: x%y)
            }
        }
        
        if z == 0 {
            return true
        }
        if x + y < z {
            return false
        } else {
            return z % gcd(x: x, y: y) == 0
        }
    }
}