LeetCode 969. Pancake Sorting

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题目

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[1...k].

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 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:

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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.
Notice that we return an array of the chosen k values of the pancake flips.

Example 2:

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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).

思路

煎饼排序: 每次找最大的煎饼，先翻到第一个位置，再翻到最后一个位置，经过2n次，可以得到结果。

代码

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class Solution:
    def pancakeSort(self, arr: List[int]) -> List[int]:
        res, n = [], len(arr)
        for i in range(n, 0 ,-1):
            pos = arr.index(i)
            if pos == i-1:
                continue
            if pos != 0:
                res.append(pos+1)
                arr[:pos+1] = arr[:pos+1][::-1]
            res.append(i)
            arr[:i] = arr[:i][::-1]
        return res