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:
Choose an integer k where 1 <= k <= arr.length.
Reverse the sub-array arr[1...k].
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 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.
Notice that we return an array of the chosen k values of the pancake flips.
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
All integers in arr are unique (i.e. arr is a permutation of the integers from 1 to arr.length).
Solution:
class Solution {
public List<Integer> pancakeSort(int[] arr) {
List<Integer> result = new ArrayList();
for (int largest = arr.length; largest > 0; largest --) {
int index = find(arr, largest);
reverse(arr, index);
reverse(arr, largest - 1);
result.add(index + 1);
result.add(largest);
}
return result;
}
private int find(int[] arr, int target) {
for (int i = 0; i < arr.length; i ++) {
if (arr[i] == target) {
return i;
}
}
return -1;
}
private void reverse(int[] arr, int right) {
int left = 0;
while (left < right) {
int temp = arr[left];
arr[left ++] = arr[right];
arr[right --] = temp;
}
}
}