【leetcode】969. Pancake Sorting

题目以下：

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]

解题思路：本题没有要求求出最少的操做次数，并且规定了操做上限是 10 * A.length。个人方法是首先把最大的数移到最后，这里只须要两步，第一是找出最大的数的位置，而且把这一段反转，这样的话最大数就在第一位了，接下来逆转整个数组，最大数就被翻转到最后。接下来是次大数，方法也同样，只不过在操做的过程当中不处理数组最后一个元素（即已经放好位置的最大数）。这种方法理论上的最大操做次数是 2 * A.length。

代码以下：

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