[Swift]LeetCode808. 分湯 | Soup Servings

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There are two types of soup: type A and type B. Initially we have N ml of each type of soup. There are four kinds of operations:

1. Serve 100 ml of soup A and 0 ml of soup B
2. Serve 75 ml of soup A and 25 ml of soup B
3. Serve 50 ml of soup A and 50 ml of soup B
4. Serve 25 ml of soup A and 75 ml of soup B

When we serve some soup, we give it to someone and we no longer have it.  Each turn, we will choose from the four operations with equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as we can.  We stop once we no longer have some quantity of both types of soup.

Note that we do not have the operation where all 100 ml's of soup B are used first.  

Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. 

Example:
Input: N = 50
Output: 0.625
Explanation: 
If we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.

Notes:

• 0 <= N <= 10^9. 
• Answers within 10^-6 of the true value will be accepted as correct.

---

有 A 和 B 兩種類型的湯。一開始每種類型的湯有 N 毫升。有四種分配操做：

1. 提供 100ml 的湯A 和 0ml 的湯B。
2. 提供 75ml 的湯A 和 25ml 的湯B。
3. 提供 50ml 的湯A 和 50ml 的湯B。
4. 提供 25ml 的湯A 和 75ml 的湯B。

當咱們把湯分配給某人以後，湯就沒有了。每一個回合，咱們將從四種機率同爲0.25的操做中進行分配選擇。若是湯的剩餘量不足以完成某次操做，咱們將盡量分配。當兩種類型的湯都分配完時，中止操做。

注意不存在先分配100 ml湯B的操做。

須要返回的值： 湯A先分配完的機率 + 湯A和湯B同時分配完的機率 / 2。

示例:
輸入: N = 50
輸出: 0.625
解釋:
若是咱們選擇前兩個操做，A將首先變爲空。對於第三個操做，A和B會同時變爲空。對於第四個操做，B將首先變爲空。
因此A變爲空的總機率加上A和B同時變爲空的機率的一半是 0.25 *(1 + 1 + 0.5 + 0)= 0.625。

註釋:

• 0 <= N <= 10^9。

• 返回值在 10^-6 的範圍將被認爲是正確的。

---

Runtime: 12 ms

Memory Usage: 19 MB

 1 class Solution {
 2     var m:[String:Double] = [String:Double]()
 3     func soupServings(_ N: Int) -> Double {
 4         return N >= 4800 ? 1.0 : f(N, N)
 5     }
 6     
 7     func f(_ a:Int,_ b:Int) ->Double
 8     {
 9         if a <= 0 && b <= 0 {return 0.5}
10         if a <= 0 {return 1.0}
11         if b <= 0 {return 0}
12         var spoon:String = String(a) + ":" + String(b)
13         if m[spoon] == nil
14         {
15             m[spoon] = 0.25 * (f(a - 100, b) + f(a - 75, b - 25) + f(a - 50, b - 50) + f(a - 25, b - 75))
16         }
17         return m[spoon]!
18     }
19 }

---

72ms

 1 class Solution {
 2     var mark = [String : Double]()
 3     func soupServings(_ N: Int) -> Double {
 4         return N > 4800 ? 1 : distributeSoup(N, N)
 5     }
 6     
 7     func distributeSoup(_ a: Int, _ b: Int) -> Double {
 8         let key = "\(a),\(b)"
 9         if let rate = mark[key] {
10             return rate
11         }
12         
13         if a <= 0 {
14             if b > 0 {
15                 return 1
16             } else {
17                 return 0.5
18             }
19         } else {
20             if b <= 0 {
21                 return 0
22             }
23         }
24         
25         let distribute = distributeSoup(a - 100, b)
26             + distributeSoup(a - 75, b - 25)
27             + distributeSoup(a - 50, b - 50)
28             + distributeSoup(a - 25, b - 75)
29         mark[key] = distribute * 0.25
30         return mark[key] ?? 0
31     }
32 }