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, 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).
class Solution {
private int findMaxIndex(int[] arr,int index){
int result=Integer.MIN_VALUE;
for(int i=0;i<arr.length;i++){
if(arr[i]==index){
result=i;
}
}
return result;
}
private void flipIt(int[] arr,int index){
int start=0;
while(start<index){
int temp=arr[start];
arr[start]=arr[index];
arr[index]=temp;
start++;
index--;
}
}
public List<Integer> pancakeSort(int[] A) {
List<Integer> list=new ArrayList<Integer>();
int max_index,len=A.length;
while(len!=1){
max_index=findMaxIndex(A,len);
flipIt(A,max_index);
flipIt(A,len-1);
list.add(max_index+1);
list.add(len);
len--;
}
return list;
}
}