題目如下：

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]

解題思路：本題沒有要求求出最少的操作次數，而且規定了操作上限是 10 * A.length。我的方法是首先把最大的數移到最後，這裡只需要兩步，第一是找出最大的數的位置，並且把這一段反轉，這樣的話最大數就在第一位了，接下來逆轉整個陣列，最大數就被翻轉到最後。接下來是次大數，方法也一樣，只不過在操作的過程中不處理陣列最後一個元素（即已經放好位置的最大數）。這種方法理論上的最大操作次數是 2 * A.length。

程式碼如下：

class Solution(object):
    def pancakeSort(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        bound = len(A)
        sa = sorted(A)
        end = len(A)
        res = []
        while A != sa:
            inx = A.index(bound)
            if inx != 0:
                A = A[0:inx+1][::-1] + A[inx+1:]
                res.append(inx+1)
            A = A[:end][::-1] + A[end:]
            res.append(end)
            end -= 1
            bound -= 1
        return res