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
where1 <= 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 from1
toarr.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