# 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 <= A.length <= 100

A[i] is a permutation of [1, 2, ..., A.length]

**Explanation**Find the index

`i`

of the next maximum number `x`

. Reverse `i + 1`

numbers, so that the `x`

will be at `A[0]`

Reverse `x`

numbers, so that `x`

will be at `A[x - 1]`

. Repeat this process `N`

times.**Update:**Actually, I didn't use the condition permutation of

`[1,2,..., A.length]`

. I searched in the descending order of `A`

.**Time Complexity**: O(N^2)

class Solution {

public List<Integer> pancakeSort(int[] A) {

List<Integer> res = new ArrayList<>();

for (int x = A.length, i; x > 0; --x) {

for (i = 0; A[i] != x; ++i);

reverse(A, i + 1); // 把最大的值移到A[0]

res.add(i + 1);

reverse(A, x); // 把最大的移到A[end]

res.add(x);

}

return res;

}

public void reverse(int[] A, int k) {

for (int i = 0, j = k - 1; i < j; i++, j--) {

int tmp = A[i];

A[i] = A[j];

A[j] = tmp;

}

}

}

**Complexity Analysis**

- Time Complexity: O(N^2), where N
*N*is the length of`A`

. - Space Complexity: O(N).

class Solution {

public List<Integer> pancakeSort(int[] A) {

List<Integer> ans = new ArrayList();

int N = A.length;

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]);

for (int i: B) {

for (int f: ans) {

if (i <= f) {

i = f + 1 - i;

}

}

ans.add(i);

ans.add(N--);

}

return ans;

}

}

Last modified 3yr ago