Given an array A, we can perform a pancake flip: We choose some positive integer k <= A.length, then reverse the order of the first k elements of A.  We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A

Return the k-values corresponding to a sequence of pancake flips that sort A.  Any valid answer that sorts the array within 10 * A.length flips will be judged as correct.

Example 1:

Input: [3,2,4,1]
Output: [4,2,4,3]
Explanation: 
We perform 4 pancake flips, with k values 4, 2, 4, and 3.
Starting state: A = [3, 2, 4, 1]
After 1st flip (k=4): A = [1, 4, 2, 3]
After 2nd flip (k=2): A = [4, 1, 2, 3]
After 3rd flip (k=4): A = [3, 2, 1, 4]
After 4th flip (k=3): A = [1, 2, 3, 4], which is sorted. 
 

Example 2:

Input: [1,2,3]
Output: []
Explanation: The input is already sorted, so there is no need to flip anything.
Note that other answers, such as [3, 3], would also be accepted.

Note:

1. 1 <= A.length <= 100
2. A[i] is a permutation of [1, 2, ..., A.length]

給定陣列 A

返回能使 A 排序的煎餅翻轉操作所對應的 k 值序列。任何將陣列排序且翻轉次數在 10 * A.length 範圍內的有效答案都將被判斷為正確。

示例 1：

輸入：[3,2,4,1]
輸出：[4,2,4,3]
解釋：
我們執行 4 次煎餅翻轉，k 值分別為 4，2，4，和 3。
初始狀態 A = [3, 2, 4, 1]
第一次翻轉後 (k=4): A = [1, 4, 2, 3]
第二次翻轉後 (k=2): A = [4, 1, 2, 3]
第三次翻轉後 (k=4): A = [3, 2, 1, 4]
第四次翻轉後 (k=3): A = [1, 2, 3, 4]，此時已完成排序。 

示例 2：

輸入：[1,2,3]
輸出：[]
解釋：
輸入已經排序，因此不需要翻轉任何內容。
請注意，其他可能的答案，如[3，3]，也將被接受。

提示：

1. 1 <= A.length <= 100
2. A[i] 是 [1, 2, ..., A.length] 的排列

40ms

 1 class Solution {
 2     func pancakeSort(_ A: [Int]) -> [Int] {
 3         var A = A
 4         var n:Int = A.count
 5         var ans:[Int] = [Int]()
 6         for i in (0...(n - 1)).reversed()
 7         {
 8             var j:Int = 0
 9             while(A[j] != i+1)
10             {
11                 j += 1
12             }
13             ans.append(j + 1)
14             ans.append(i + 1)
15             var newA:[Int] = [Int](repeating:0,count:n)
16             for k in 0..<n
17             {
18                 newA[k] = A[k]
19             }
20             for k in 0...j
21             {
22                 newA[k] = A[j-k]
23             }
24             A = newA
25             newA = [Int](repeating:0,count:n)
26             for k in 0..<n
27             {
28                 newA[k] = A[k]
29             }
30             for k in 0...i
31             {
32                 newA[k] = A[i-k]
33             }
34             A = newA
35         }
36         return ans
37     }
38 }