[Swift]LeetCode955. 删列造序 II | Delete Columns to Make Sorted II

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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 an array A = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then the final array after deletions is ["bef","vyz"].

Suppose we chose a set of deletion indices D such that after deletions, the final array has its elements in lexicographic order (A[0] <= A[1] <= A[2] ... <= A[A.length - 1]).

Return the minimum possible value of D.length.

Example 1:

Input: ["ca","bb","ac"]
Output: 1 Explanation: After deleting the first column, A = ["a", "b", "c"]. Now A is in lexicographic order (ie. A[0] <= A[1] <= A[2]). We require at least 1 deletion since initially A was not in lexicographic order, so the answer is 1. 

Example 2:

Input: ["xc","yb","za"]
Output: 0
Explanation: 
A is already in lexicographic order, so we don't need to delete anything.
Note that the rows of A are not necessarily in lexicographic order:
ie. it is NOT necessarily true that (A[0][0] <= A[0][1] <= ...)

Example 3:

Input: ["zyx","wvu","tsr"]
Output: 3 Explanation: We have to delete every column.

Note:

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

---

给定由 N 个小写字母字符串组成的数组 A，其中每一个字符串长度相等。

选取一个删除索引序列，对于 A 中的每一个字符串，删除对应每一个索引处的字符。

好比，有 A = ["abcdef", "uvwxyz"]，删除索引序列 {0, 2, 3}，删除后 A 为["bef", "vyz"]。

假设，咱们选择了一组删除索引 D，那么在执行删除操做以后，最终获得的数组的元素是按 字典序（A[0] <= A[1] <= A[2] ... <= A[A.length - 1]）排列的，而后请你返回 D.length 的最小可能值。

示例 1：

输入：["ca","bb","ac"]
输出：1
解释： 
删除第一列后，A = ["a", "b", "c"]。
如今 A 中元素是按字典排列的 (即，A[0] <= A[1] <= A[2])。
咱们至少须要进行 1 次删除，由于最初 A 不是按字典序排列的，因此答案是 1。

示例 2：

输入：["xc","yb","za"]
输出：0
解释：
A 的列已是按字典序排列了，因此咱们不须要删除任何东西。
注意 A 的行不须要按字典序排列。
也就是说，A[0][0] <= A[0][1] <= ... 不必定成立。

示例 3：

输入：["zyx","wvu","tsr"]
输出：3
解释：
咱们必须删掉每一列。

提示：

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

---

72ms

 1 class Solution {
 2     func minDeletionSize(_ A: [String]) -> Int {
 3         var n:Int = A.count
 4         var ss:[String] = [String](repeating:"",count:n)
 5         var ret:Int = 0
 6         outer:
 7         for i in 0..<A[0].count
 8         {
 9             for j in 0..<(n - 1)
10             {
11                 if ss[j] == ss[j+1] && A[j][i] > A[j+1][i]
12                 {
13                     ret += 1
14                     continue outer
15                 }
16             }
17             for j in 0..<n
18             {
19                 ss[j].append(A[j][i])
20             }
21         }
22         
23         return ret                  
24     }
25 }
26     
27 extension String {        
28     //subscript函数能够检索数组中的值
29     //直接按照索引方式截取指定索引的字符
30     subscript (_ i: Int) -> Character {
31         //读取字符
32         get {return self[index(startIndex, offsetBy: i)]}
33         
34     }
35 }

---

100ms

 1 class Solution {
 2     func minDeletionSize(_ A: [String]) -> Int {
 3         let R = A.count
 4         guard A.count > 1 else {
 5             return 0
 6         }
 7         let C = A[0].count
 8         var del = [Int](repeating: 0, count: C)
 9         var pc = [Int](repeating: 0, count: R)
10         for c in 0 ..< C {
11             var npc = [Int](repeating: 0, count: R)
12             for r in 1 ..< R {
13                 let ic = A[r].index(A[r].startIndex, offsetBy: c)
14                 let v = A[r][ic] < A[r-1][ic] ? 1 : A[r][ic] == A[r-1][ic] ? 0 : -1
15                 npc[r] = pc[r] == -1 ? -1 : v
16                 //print(A[r][ic], A[r-1][ic], v)
17                 if npc[r] > 0 {
18                     del[c] = 1
19                 }                    
20             }
21             //print(pc, npc, del)
22             if del[c] == 0 {
23                 pc = npc
24             }
25         }
26         return del.reduce(0, +)
27     }
28 }