题目如下:
Given an array
A
, we can perform a pancake flip: We choose some positive integerk <= A.length
, then reverse the order of the first k elements ofA
. We want to perform zero or more pancake flips (doing them one after another in succession) to sort the arrayA
.Return the k-values corresponding to a sequence of pancake flips that sort
A
. Any valid answer that sorts the array within10 * 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 <= A.length <= 100
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作次数,而且规定了操作上限是 10 * A.length。我的方法是首先把最大的数移到最后,这里只需要两步,第一是找出最大的数的位置,并且把这一段反转,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,方法也一样,只不过在操作的过程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作次数是 2 * A.length。
代码如下:
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
原文地址:https://www.cnblogs.com/seyjs/p/10233848.html
时间: 2024-10-02 03:29:54