# [969. Pancake Sorting (Medium)](https://leetcode.com/problems/pancake-sorting/)

<p>Given an array <code>A</code>, we can perform a&nbsp;<em>pancake flip</em>:&nbsp;We choose some positive integer&nbsp;<code><strong>k</strong> &lt;= A.length</code>, then reverse the order of the first <strong>k</strong> elements of <code>A</code>.&nbsp; We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array <code>A</code>.</p>

<p>Return the k-values corresponding to a sequence of pancake flips that sort <code>A</code>.&nbsp; Any&nbsp;valid answer that sorts the array within <code>10 * A.length</code> flips will be judged as correct.</p>

<p>&nbsp;</p>

<p><strong>Example 1:</strong></p>

<pre><strong>Input: </strong><span id="example-input-1-1">[3,2,4,1]</span>
<strong>Output: </strong><span id="example-output-1">[4,2,4,3]</span>
<strong>Explanation: </strong>
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. 
</pre>

<div>
<p><strong>Example 2:</strong></p>

<pre><strong>Input: </strong><span id="example-input-2-1">[1,2,3]</span>
<strong>Output: </strong><span id="example-output-2">[]</span>
<strong>Explanation: </strong>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.
</pre>

<p>&nbsp;</p>
</div>

<p><strong>Note:</strong></p>

<ol>
	<li><code>1 &lt;= A.length &lt;= 100</code></li>
	<li><code>A[i]</code> is a permutation of <code>[1, 2, ..., A.length]</code></li>
</ol>


**Companies**:  
[Square](https://leetcode.com/company/square)

**Related Topics**:  
[Array](https://leetcode.com/tag/array/), [Sort](https://leetcode.com/tag/sort/)

## Solution 1.

```cpp
// OJ: https://leetcode.com/problems/pancake-sorting
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    vector<int> pancakeSort(vector<int>& A) {
        vector<int> ans;
        for (int i = A.size(); i > 0; --i) {
            int j = i - 1;
            for (; j >= 0 && A[j] != i; --j);
            reverse(A.begin(), A.begin() + j + 1);
            ans.push_back(j + 1);
            reverse(A.begin(), A.begin() + i);
            ans.push_back(i);
        }
        return ans;
    }
};
```