# LeetCode-969. Pancake Sorting

## LeetCode-969. Pancake Sorting

Posted by Jae on October 31, 2019

### 1、题目

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]


### 3 实现

class Solution {
public List<Integer> pancakeSort(int[] A) {
int N = A.length;
List<Integer> ans = new ArrayList<>();

Integer[] B = new Integer[N];
for (int i = 0; i < N; i++){
B[i] = i+1;
}
Arrays.sort(B, (i, j) -> A[j-1]-A[i-1]);
// System.out.println(Arrays.toString(B));
for (int i : B){
for (int move : ans){
if (i <= move){
i = move + 1 - i;
}
}