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 kelements 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.lengthflips 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]

思路：從N遍歷到1，每次先解決剩下的最大的數，比如遍歷到10（10在第5位上），也就是說10-N已經在正確的位置上，1-10還沒有。那就先flip 1-5把10放到第一位，接著flip 1-10把10放到正確的位置上。

保證2N次交換一定可以

class Solution:
    def pancakeSort(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        if tuple(A)==tuple(sorted(A)): return []
        d={v:i+1 for i,v in enumerate(A)}
        res = []
        for i in range(len(A),0,-1):
            res.append(d[i])
            res.append(i)
            
            d2={}
            for s in d:
                if s>i: continue
                pos=d[s]
                if pos<=d[i]:
                    pos=d[i]+1-pos
                pos=i+1-pos
                d2[s] = pos
            d=d2
        
        return res