[Swift Weekly Contest 111]LeetCode944. 删除列以使之有序 | Delete Columns to Make Sorted

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

现在，我们可以选择任何一组删除索引，对于每个字符串，我们将删除这些索引中的所有字符。

举个例子，如果字符串为 "abcdef"，且删除索引是 {0, 2, 3}，那么删除之后的最终字符串为 "bef"。

假设我们选择了一组删除索引 D，在执行删除操作之后，A 中剩余的每一列都是有序的。

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

返回 D.length 的最小可能值。

示例 1：

输入：["cba","daf","ghi"]
输出：1

示例 2：

输入：["a","b"]
输出：0

示例 3：

输入：["zyx","wvu","tsr"]
输出：3

提示：

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

超出时间限制

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

原文地址：https://www.cnblogs.com/strengthen/p/9977742.html