Given an array of integers A, We need to sort the array performing a series of pancake flips.

In one pancake flip we do the following steps:

Choose an integer k where 0 <= k < A.length.
Reverse the sub-array A[0...k]. For example, if A = [3,2,1,4] and we performed a pancake flip choosing k = 2</font>**, we reverse the sub-array **[3,2,1], so **A = [1,2,3,4]** after the pancake flip at **k = 2**.

Return an array of the k-values of the pancake flips that should be performed in order to sort A. Any valid answer that sorts the array within 10 * A.length flips will be judged as correct.

Example 1:

Input: A = [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.
Notice that we return an array of the chosen k values of the pancake flips.

Example 2:

Input: A = [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 <= A.length <= 100
1 <= A[i] <= A.length
All integers in A are unique (i.e. A is a permutation of the integers from 1 to A.length).

solution

Time complexity : O(n^2) Space complexity : O(1)

class Solution {
public:
    vector<int> pancakeSort(vector<int>& A) {
        int n = A.size();
        vector<int> res = {};
        for (int i=0; i<n-1; ++i) {
            int pos = max_element(A.begin(), A.end()-i) - A.begin();
            res.push_back(pos+1);
            reverse(A.begin(), A.begin()+(pos+1));
            res.push_back(n-i);
            reverse(A.begin(), A.begin()+(n-i));
        }
        return res;
    }
};

從右至左，數值大至小處理。
每次處理一個數時，找到最大值，將它移到最左邊，再將它移到最右邊。