LC 969. Pancake Sorting

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]

Runtime: 14 ms

class Solution {
  static public List<Integer> pancakeSort(int[] A) {
    List<Integer> ret = new ArrayList<>();
    for(int x = A.length,i=; x>= ; x--){
      //System.out.print(x);
      for(i=; A[i] != x; i++);
      reverse(A, i);
      ret.add(i+);
      reverse(A, x-);
      ret.add(x);
    }
    return ret;
  }
  static void reverse(int[] A, int pos){
    for(int i=,j=pos; i<j; i++,j--){
      int tmp = A[i];
      A[i] = A[j];
      A[j] = tmp;
    }
  }
}