[Swift]LeetCode433. 最小基因變化 | Minimum Genetic Mutation

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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A gene string can be represented by an 8-character long string, with choices from "A", "C", "G", "T".

Suppose we need to investigate about a mutation (mutation from "start" to "end"), where ONE mutation is defined as ONE single character changed in the gene string.

For example, "AACCGGTT" -> "AACCGGTA" is 1 mutation.

Also, there is a given gene "bank", which records all the valid gene mutations. A gene must be in the bank to make it a valid gene string.

Now, given 3 things - start, end, bank, your task is to determine what is the minimum number of mutations needed to mutate from "start" to "end". If there is no such a mutation, return -1.

Note:

1. Starting point is assumed to be valid, so it might not be included in the bank.
2. If multiple mutations are needed, all mutations during in the sequence must be valid.
3. You may assume start and end string is not the same. 

Example 1:

start: "AACCGGTT"
end:   "AACCGGTA"
bank: ["AACCGGTA"]

return: 1 

Example 2:

start: "AACCGGTT"
end:   "AAACGGTA"
bank: ["AACCGGTA", "AACCGCTA", "AAACGGTA"]

return: 2 

Example 3:

start: "AAAAACCC"
end:   "AACCCCCC"
bank: ["AAAACCCC", "AAACCCCC", "AACCCCCC"]

return: 3

---

一條基因序列由一個帶有8個字符的字符串表示，其中每一個字符都屬於 "A", "C", "G", "T"中的任意一個。

假設咱們要調查一個基因序列的變化。一次基因變化意味着這個基因序列中的一個字符發生了變化。

例如，基因序列由"AACCGGTT" 變化至 "AACCGGTA" 即發生了一次基因變化。

與此同時，每一次基因變化的結果，都須要是一個合法的基因串，即該結果屬於一個基因庫。

如今給定3個參數 — start, end, bank，分別表明起始基因序列，目標基因序列及基因庫，請找出可以使起始基因序列變化爲目標基因序列所需的最少變化次數。若是沒法實現目標變化，請返回 -1。

注意:

1. 起始基因序列默認是合法的，可是它並不必定會出如今基因庫中。
2. 全部的目標基因序列必須是合法的。
3. 假定起始基因序列與目標基因序列是不同的。

示例 1:

start: "AACCGGTT"
end:   "AACCGGTA"
bank: ["AACCGGTA"]

返回值: 1

示例 2:

start: "AACCGGTT"
end:   "AAACGGTA"
bank: ["AACCGGTA", "AACCGCTA", "AAACGGTA"]

返回值: 2

示例 3:

start: "AAAAACCC"
end:   "AACCCCCC"
bank: ["AAAACCCC", "AAACCCCC", "AACCCCCC"]

返回值: 3

---

8ms

 1 class Solution {
 2     func minMutation(_ start: String, _ end: String, _ bank: [String]) -> Int {
 3         var bank = bank
 4         var explored:[Bool] = [Bool](repeating:false,count:bank.count)
 5         if bank.isEmpty {return -1}
 6         return minMutation(&explored, start, end, &bank)
 7     }
 8     
 9     func minMutation(_ explored:inout [Bool],_ start:String,_ end:String,_ bank:inout[String]) -> Int
10     {
11         if start == end {return 0}
12         var step:Int = bank.count + 1
13         for i in 0..<bank.count
14         {
15             if diffOne(start, bank[i]) && !explored[i]
16             {
17                 explored[i] = true
18                 var temp:Int = minMutation(&explored, bank[i], end, &bank)
19                 if temp != -1
20                 {
21                     step = min(step, temp)
22                 }
23                 explored[i] = false
24             }            
25         }
26         return step == bank.count + 1 ? -1 : 1 + step
27     }
28     
29     func diffOne(_ s1:String,_ s2:String) -> Bool
30     {
31         var count:Int = 0
32         for i in 0..<s1.count
33         {
34             if s1[i] != s2[i]
35             {
36                 count += 1
37             }
38             if count >= 2
39             {
40                 return false
41             }
42         }
43         return count == 1
44     }
45 }
46 
47 extension String {        
48     //subscript函數能夠檢索數組中的值
49     //直接按照索引方式截取指定索引的字符
50     subscript (_ i: Int) -> Character {
51         //讀取字符
52         get {return self[index(startIndex, offsetBy: i)]}
53         
54     }
55 }

---

10ms

 1 class Solution {
 2     func minMutation(_ start: String, _ end: String, _ bank: [String]) -> Int {
 3         let geneChoice = [Character("A"), Character("C"), Character("G"), Character("T")]
 4         var bankArray = Set(bank.map { Array($0) })
 5         var startArray = Array(start)
 6         var endArray = Array(end)
 7         var queue = [startArray]
 8         var mutationCount = 0
 9         while queue.count > 0 {
10             var nextQueue = [[Character]]()
11             for i in 0 ..< queue.count {
12                 var curr = queue[i]
13                 if curr == endArray {
14                     return mutationCount
15                 }
16                 for j in 0 ..< 8 {
17                     var c = curr[j]
18                     for k in 0 ..< geneChoice.count {
19                         if c != geneChoice[k] {
20                             curr[j] = geneChoice[k]
21                             if bankArray.contains(curr) {
22                                 nextQueue.append(curr)
23                                 bankArray.remove(curr)
24                             }
25                         }
26                     }
27                     curr[j] = c
28                 }
29             }
30             queue = nextQueue
31             mutationCount += 1
32         }
33         return -1
34     }
35 }

---

12ms

 1 class Solution {
 2       func minMutation(_ start: String, _ end: String, _ bank: [String]) -> Int {
 3         if !bank.contains(end){
 4             return -1
 5         }
 6         if start == end{
 7             return 0
 8         }
 9         var max :Int = 0
10         
11         var arr = [String]()
12         var vists = [String]()
13         var temp = [String]()
14         arr.append(start)
15         while arr.count > 0 {
16       
17             for j in 0..<arr.count{
18                 for k in 0..<bank.count{
19                     if vists.contains(bank[k]){continue}
20                     var dif :Int = 0
21                     for  i in 0..<8{
22                         let s = arr[j].index(arr[j].startIndex, offsetBy: i)
23                         let b = bank[k].index(bank[k].startIndex, offsetBy: i)
24                         if arr[j][s] != bank[k][b] {
25                             dif = dif + 1
26                         }
27                         if(i == 7  && dif == 1){
28                             if bank[k] == end {return max + 1}
29                             temp.append(bank[k])
30                             vists.append(bank[k])
31                         }
32                     }
33                     
34                     
35                 }
36             }
37             if temp.count == 0 {return -1}else{ max = max + 1 }
38             arr = temp;temp = [];
39         }
40         return max
41     }
42 }