难度: Medium
原题连接
内容描述
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 <= A.length <= 100
A[i] is a permutation of [1, 2, ..., A.length]
思路 1 - 时间复杂度: O(N^2)- 空间复杂度: O(1)******
服了，题目也没说是个答案就可以啊，可以自己构造出来，选择排序的思想
- 找出当前最大的那个数len(A)，先reverse到第一位，再reverse到最后一位
- 找出当前最大的那个数len(A)-1，先reverse到第一位，再reverse到最后一位
- ......
- ......
- ......
- 找出当前最大的那个数1，先reverse到第一位，再reverse到最后一位
class Solution:
def pancakeSort(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
if A == sorted(A):
return []
res, k = [], 0
for i in range(len(A), 0, -1): # i 是当前还未排好序的最大的数字
idx = 0
for j, num in enumerate(A):
if num == i:
idx = j
res.append(idx + 1) # 先reverse到第一位
A = A[:idx+1][::-1] + A[idx+1:]
res.append(len(A) - k) # 再reverse到最后一位
A = A[:len(A)-k][::-1] + A[len(A)-k:]
k += 1
return res