# 969. Pancake Sorting 煎饼排序

@TOC

## # 题目描述

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. 1 <= A.length <= 100
2. A[i] is a permutation of [1, 2, ..., A.length]

## # 解题方法

### # 模拟法

C++代码如下：

``````class Solution {
public:
vector<int> pancakeSort(vector<int>& A) {
vector<int> res;
const int N = A.size();
int pos, i;
for (pos = N; pos > 0; pos--) {
for (i = 0; A[i] != pos; i ++);
reverse(A.begin(), A.begin() + i + 1);
if (i != 0)
res.push_back(i + 1);
reverse(A.begin(), A.begin() + pos);
res.push_back(pos);
}
return res;
}
};
``````

python代码简洁一点，可以直接使用Index函数找到x，然后再翻转，但是这个翻转操作还不如C++的reverse函数来的简单。

``````class Solution(object):
def pancakeSort(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
N = len(A)
res = []
for x in range(N, 0, -1):
i = A.index(x)
res.extend([i + 1, x])
A = A[:i:-1] + A[:i]
return res
``````

``````class Solution(object):
def pancakeSort(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
N = len(A)
res = []
for x in range(N, 0, -1):
i = A.index(x)
if i != 0:
res.append(i + 1)
res.append(x)
self.reverse(A, 0, i + 1);
self.reverse(A, 0, x);
print(A)
return res

#[start, end)
def reverse(self, A, start, end):
A[start:end] = A[start:end][::-1]
``````

## # 日期

2019 年 1 月 6 日 —— 打球打的腰酸背痛