[Swift]LeetCode944. 刪除列以使之有序 | Delete Columns to Make Sorted

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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We are given an array A of N lowercase letter strings, all of the same length.

Now, we may choose any set of deletion indices, and for each string, we delete all the characters in those indices.

For example, if we have a string "abcdef" and deletion indices {0, 2, 3}, then the final string after deletion is "bef".

Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.

Formally, the c-th column is [A[0][c], A[1][c], ..., A[A.length-1][c]]

Return the minimum possible value of D.length.

 Example 1:

Input: ["cba","daf","ghi"]
Output: 1 

Example 2:

Input: ["a","b"]
Output: 0 

Example 3:

Input: ["zyx","wvu","tsr"]
Output: 3

Note:

1. 1 <= A.length <= 100
2. 1 <= A[i].length <= 1000

---

給出由 N 個小寫字母串組成的數組 A，全部小寫字母串的長度都相同。

如今，咱們能夠選擇任何一組刪除索引，對於每一個字符串，咱們將刪除這些索引中的全部字符。

舉個例子，若是字符串爲 "abcdef"，且刪除索引是 {0, 2, 3}，那麼刪除以後的最終字符串爲 "bef"。

假設咱們選擇了一組刪除索引 D，在執行刪除操做以後，A 中剩餘的每一列都是有序的。

形式上，第 c 列爲 [A[0][c], A[1][c], ..., A[A.length-1][c]]

返回 D.length 的最小可能值。

 示例 1：

輸入：["cba","daf","ghi"]
輸出：1

示例 2：

輸入：["a","b"]
輸出：0

示例 3：

輸入：["zyx","wvu","tsr"]
輸出：3

提示：

1. 1 <= A.length <= 100
2. 1 <= A[i].length <= 1000

---

232ms 

 1 class Solution {
 2   func minDeletionSize(_ A: [String]) -> Int {
 3     var dl = 0
 4     var Ab : [[UInt8]] = []
 5     for var s in A {
 6       Ab.append(Array<UInt8>(s.utf8))
 7     }
 8     for var i in 0..<Ab[0].count {
 9       for var j in 0..<A.count-1 {
10         if Ab[j][i] > Ab[j+1][i] {
11           dl += 1
12           break
13         }
14       } 
15     }
16     return dl
17   }
18 }

---

264ms

 1 class Solution {
 2     func minDeletionSize(_ A: [String]) -> Int {
 3         guard A.count > 1 else { return 0 }
 4         var minSet = Set<Int>()
 5         for i in 0..<A.count-1 {
 6             let strArr = Array(A[i])
 7             let strArr2 = Array(A[i+1])
 8             for k in 0..<strArr.count {
 9                 if strArr[k] > strArr2[k] {
10                     minSet.insert(k)
11                 }
12             }
13         }
14         return minSet.count
15     }
16 }

---

272ms

 1 class Solution {
 2     func minDeletionSize(_ A: [String]) -> Int {
 3         var chars: [[Character]] = []
 4         chars = A.map { Array($0) }
 5         let numColumns = A.first!.count
 6         var count = 0
 7         for i in 0..<numColumns {
 8             inner: for j in 1..<chars.count {
 9                 if chars[j][i] < chars[j - 1][i] {
10                     count += 1
11                     break inner
12                 }
13             }
14         }
15         return count
16     }
17 }

---

280ms

 1 class Solution {
 2   func minDeletionSize(_ A: [String]) -> Int {
 3     var deleteCount = 0
 4     
 5     var arr = [[Character]]()
 6     for i in 0..<A.count {
 7       arr.append(Array(A[i]))
 8     }
 9     
10     for i in 0..<arr[0].count {
11       for j in 1..<arr.count {
12         if arr[j-1][i] > arr[j][i] {
13           deleteCount += 1
14           break
15         }
16       }
17     }
18     return deleteCount
19   }
20 }

---

384ms

 1 class Solution {
 2     func minDeletionSize(_ A: [String]) -> Int {
 3        var d_size = [Int]()
 4     for index_a in A.indices {
 5         if (index_a+1) < A.count {
 6             let a_s1 = Array(A[index_a])
 7             let a_s2 = Array(A[index_a+1])
 8             for s_i in 0..<a_s1.count {
 9                 if String(a_s1[s_i]) > String(a_s2[s_i]) && !d_size.contains(s_i)  {
10                     d_size.append(s_i)
11                 }
12             }
13         }
14         
15     }
16     return d_size.count 
17     }
18 }