動機

模擬bubble sort

Problem

Given an array of integers arr, sort the array by performing a series of pancake flips.

In one pancake flip we do the following steps:

  • Choose an integer k where 1 <= k <= arr.length.
  • Reverse the sub-array arr[0...k-1] (0-indexed).

For example, if arr = [3,2,1,4] and we performed a pancake flip choosing k = 3, we reverse the sub-array [3,2,1], so arr = [1,2,3,4] after the pancake flip at k = 3.

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

 

Example 1:

Input: arr = [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: arr = [3, 2, 4, 1]After 1st flip (k = 4): arr = [1, 4, 2, 3]After 2nd flip (k = 2): arr = [4, 1, 2, 3]After 3rd flip (k = 4): arr = [3, 2, 1, 4]After 4th flip (k = 3): arr = [1, 2, 3, 4], which is sorted.

Example 2:

Input: arr = [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.

 

Constraints:

  • 1 <= arr.length <= 100
  • 1 <= arr[i] <= arr.length
  • All integers in arr are unique (i.e. arr is a permutation of the integers from 1 to arr.length).

Sol

class Solution:
    def pancakeSort(self, arr: List[int]) -> List[int]:
        ret = []
        done = 0
        while done < len(arr):
            i = max(range(len(arr)-done), key=arr.__getitem__)
            if i+1 != len(arr)-done:
                ret += i+1,
                arr[:i+1] = arr[:i+1][::-1]
                ret += len(arr)-done,
                arr[:len(arr)-done] = arr[:len(arr)-done][::-1]
            done+=1
        return ret