[Swift]LeetCode458. 可怜的小猪 | Poor Pigs

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There are 1000 buckets, one and only one of them contains poison, the rest are filled with water. They all look the same. If a pig drinks that poison it will die within 15 minutes. What is the minimum amount of pigs you need to figure out which bucket contains the poison within one hour.

Answer this question, and write an algorithm for the follow-up general case.

Follow-up:

If there are n buckets and a pig drinking poison will die within m minutes, how many pigs (x) you need to figure out the "poison" bucket within p minutes? There is exact one bucket with poison.

---

有1000只水桶，其中有且只有一桶装的含有毒药，其余装的都是水。它们从外观看起来都一样。如果小猪喝了毒药，它会在15分钟内死去。

问题来了，如果需要你在一小时内，弄清楚哪只水桶含有毒药，你最少需要多少只猪？

回答这个问题，并为下列的进阶问题编写一个通用算法。

进阶:

假设有 n 只水桶，猪饮水中毒后会在 m 分钟内死亡，你需要多少猪（x）就能在 p 分钟内找出“有毒”水桶？n只水桶里有且仅有一只有毒的桶。

---

 8ms

class Solution {
    func poorPigs(_ buckets: Int, _ minutesToDie: Int, _ minutesToTest: Int) -> Int {
        if buckets == 1 {return 0}
        //每头小猪最多可测试的水桶数
        var  times = minutesToTest / minutesToDie + 1    
        var num:Int = 0
        //假设5桶水，一头小猪一个小时内检测四个桶
        //如果没死，则是最后一桶有毒。
        //每增加一头小猪，能检测的桶数乘5
        while(pow(Double(times),Double(num)) < Double(buckets))
        {           
            num += 1
        }
        return num
    }
}

---

8ms

class Solution {
    func poorPigs(_ buckets: Int, _ minutesToDie: Int, _ minutesToTest: Int) -> Int {
        let a: Double = log(Double(minutesToTest) / Double(minutesToDie) + 1)
        let pigs: Double = log(Double(buckets)) / a
        return Int(pigs.rounded(.up))
    }
}

---

8ms

class Solution {
    func poorPigs(_ buckets: Int, _ minutesToDie: Int, _ minutesToTest: Int) -> Int {
        var pigs = 0.0
        while pow(Double(minutesToTest / minutesToDie + 1), pigs) < Double(buckets) {
            pigs += 1
        }
        return Int(pigs)
    }
}

---

12ms

class Solution {
    func poorPigs(_ buckets: Int, _ minutesToDie: Int, _ minutesToTest: Int) -> Int {
        var pigs = 0
        while Int(pow(Double((minutesToTest / minutesToDie) + 1), Double(pigs))) < buckets {
            pigs += 1
        }
        
        return pigs
    }
}