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?微信公眾號：山青詠芝（shanqingyongzhi）
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?原文地址：https://www.cnblogs.com/strengthen/p/9977742.html?
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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 <= A.length <= 100

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

---

給出由?N?個小寫字母串組成的數組?A，所有小寫字母串的長度都相同。

現在，我們可以選擇任何一組刪除索引，對于每個字符串，我們將刪除這些索引中的所有字符。

舉個例子，如果字符串為?"abcdef"，且刪除索引是?{0, 2, 3}，那么刪除之后的最終字符串為?"bef"。

假設我們選擇了一組刪除索引?D，在執行刪除操作之后，A?中剩余的每一列都是有序的。

形式上，第?c?列為?[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 <= A.length <= 100

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 }

?

轉載于:https://www.cnblogs.com/strengthen/p/9977742.html

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