Medium
Given an array of integers arr
, sort the array by performing a series of pancake flips.
In one pancake flip we do the following steps:
k
where 1 <= k <= arr.length
.arr[0...k-1]
(0-indexed).For example, if arr = [3,2,1,4]
and we performed a pancake flip choosing k = 3
, we reverse the sub-array [3,2,1]
, so arr = [1,2,3,4]
after the pancake flip at k = 3
.
Return an array of the k
-values corresponding to a sequence of pancake flips that sort arr
. Any valid answer that sorts the array within 10 * arr.length
flips will be judged as correct.
Example 1:
Input: arr = [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: arr = [3, 2, 4, 1]
After 1st flip (k = 4): arr = [1, 4, 2, 3]
After 2nd flip (k = 2): arr = [4, 1, 2, 3]
After 3rd flip (k = 4): arr = [3, 2, 1, 4]
After 4th flip (k = 3): arr = [1, 2, 3, 4], which is sorted.
Example 2:
Input: arr = [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 <= arr.length <= 100
1 <= arr[i] <= arr.length
arr
are unique (i.e. arr
is a permutation of the integers from 1
to arr.length
).import java.util.ArrayList;
import java.util.List;
public class Solution {
public List<Integer> pancakeSort(int[] arr) {
List<Integer> result = new ArrayList<>();
for (int i = arr.length; i >= 1; i--) {
int max = Integer.MIN_VALUE;
int index = 0;
for (int j = 0; j < i; j++) {
if (max < arr[j]) {
index = j + 1;
max = arr[j];
}
}
result.add(index);
reverse(arr, index - 1);
result.add(i);
reverse(arr, i - 1);
}
return result;
}
private void reverse(int[] arr, int index) {
for (int i = 0; i <= (index - 1) / 2; i++) {
int temp = arr[i];
arr[i] = arr[index - i];
arr[index - i] = temp;
}
}
}