[Swift]LeetCode887. 鸡蛋掉落 | Super Egg Drop

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You are given K eggs, and you have access to a building with N floors from 1 to N. 

Each egg is identical in function, and if an egg breaks, you cannot drop it again.

You know that there exists a floor F with 0 <= F <= N such that any egg dropped at a floor higher than F will break, and any egg dropped at or below floor F will not break.

Each move, you may take an egg (if you have an unbroken one) and drop it from any floor X (with 1 <= X <= N). 

Your goal is to know with certainty what the value of F is.

What is the minimum number of moves that you need to know with certainty what Fis, regardless of the initial value of F? 

Example 1:

Input: K = 1, N = 2
Output: 2
Explanation: 
Drop the egg from floor 1.  If it breaks, we know with certainty that F = 0.
Otherwise, drop the egg from floor 2.  If it breaks, we know with certainty that F = 1.
If it didn't break, then we know with certainty F = 2.
Hence, we needed 2 moves in the worst case to know what F is with certainty.

Example 2:

Input: K = 2, N = 6
Output: 3

Example 3:

Input: K = 3, N = 14
Output: 4 

Note:

1. 1 <= K <= 100
2. 1 <= N <= 10000

---

你将获得 K 个鸡蛋，并可以使用一栋从 1 到 N  共有 N 层楼的建筑。

每个蛋的功能都是一样的，如果一个蛋碎了，你就不能再把它掉下去。

你知道存在楼层 F ，满足 0 <= F <= N 任何从高于 F 的楼层落下的鸡蛋都会碎，从 F 楼层或比它低的楼层落下的鸡蛋都不会破。

每次移动，你可以取一个鸡蛋（如果你有完整的鸡蛋）并把它从任一楼层 X 扔下（满足 1 <= X <= N）。

你的目标是确切地知道 F 的值是多少。

无论 F 的初始值如何，你确定 F 的值的最小移动次数是多少？ 

示例 1：

输入：K = 1, N = 2
输出：2
解释：
鸡蛋从 1 楼掉落。如果它碎了，我们肯定知道 F = 0 。
否则，鸡蛋从 2 楼掉落。如果它碎了，我们肯定知道 F = 1 。
如果它没碎，那么我们肯定知道 F = 2 。
因此，在最坏的情况下我们需要移动 2 次以确定 F 是多少。

示例 2：

输入：K = 2, N = 6
输出：3

示例 3：

输入：K = 3, N = 14
输出：4 

提示：

1. 1 <= K <= 100
2. 1 <= N <= 10000

---

Runtime: 12 ms

Memory Usage: 18.7 MB

class Solution {
    func superEggDrop(_ K: Int, _ N: Int) -> Int {
        var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:K + 1),count:N + 1)
        var m:Int = 0
        while(dp[m][K] < N)
        {
            m += 1
            for k in 1...K
            {
                dp[m][k] = dp[m - 1][k - 1] + dp[m - 1][k] + 1
            }
        }
        return m
    }
}

---

Runtime: 24 ms

Memory Usage: 19 MB

class Solution {
    var mark = [String : Int]()
    func superEggDrop(_ K: Int, _ N: Int) -> Int {
        var step = 1
        while reachHeight(K, step) < N {
            step += 1
        }
        return step
    }
    
    func reachHeight(_ K: Int, _ N: Int) -> Int {
        let key = "(K),(N)"
        if let result = mark[key] {
            return result
        }
        if K == 0 || N == 0 {
            return 0
        }
        if K == 1 || N == 1 {
            return N
        }
        var height = N
        for i in 1..<N {
            height += reachHeight(K - 1, i)
        }
        mark[key] = height
        return height
    }
}