[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

---

你将得到 K 个鸡蛋，并可使用一栋从 1 到 N  共有 N 层楼的建筑。

每一个蛋的功能都是同样的，若是一个蛋碎了，你就不能再把它掉下去。

你知道存在楼层 F ，知足 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，你能够取一个鸡蛋（若是你有完整的鸡蛋）并把它从任一楼层 X 扔下（知足 1 <= X <= N）。

你的目标是确切地知道 F 的值是多少。

不管 F 的初始值如何，你肯定 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。若是它碎了，咱们确定知道 F = 0 。
不然，鸡蛋从 2 楼掉落。若是它碎了，咱们确定知道 F = 1 。
若是它没碎，那么咱们确定知道 F = 2 。
所以，在最坏的状况下咱们须要移动 2 次以肯定 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

 1 class Solution {
 2     func superEggDrop(_ K: Int, _ N: Int) -> Int {
 3         var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:K + 1),count:N + 1)
 4         var m:Int = 0
 5         while(dp[m][K] < N)
 6         {
 7             m += 1
 8             for k in 1...K
 9             {
10                 dp[m][k] = dp[m - 1][k - 1] + dp[m - 1][k] + 1
11             }
12         }
13         return m
14     }
15 }

---

Runtime: 24 ms

Memory Usage: 19 MB

 1 class Solution {
 2     var mark = [String : Int]()
 3     func superEggDrop(_ K: Int, _ N: Int) -> Int {
 4         var step = 1
 5         while reachHeight(K, step) < N {
 6             step += 1
 7         }
 8         return step
 9     }
10     
11     func reachHeight(_ K: Int, _ N: Int) -> Int {
12         let key = "\(K),\(N)"
13         if let result = mark[key] {
14             return result
15         }
16         if K == 0 || N == 0 {
17             return 0
18         }
19         if K == 1 || N == 1 {
20             return N
21         }
22         var height = N
23         for i in 1..<N {
24             height += reachHeight(K - 1, i)
25         }
26         mark[key] = height
27         return height
28     }
29 }