0969. Pancake Sorting (M)

Pancake Sorting (M)

题目

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).

---

题意

只使用翻转操作(每次可翻转[0...k])来排序一个数组。注意k实际上表示翻转的数量。

思路

先找到最大值n，执行两次翻转，先翻转到第一位，再翻转到最后一位，这样n就已经在它应在的位置上了。再找到n-1，将其翻转到倒数第二位。以此类推。

---

代码实现

Java

class Solution {
    public List<Integer> pancakeSort(int[] A) {
        List<Integer> ans = new ArrayList<>();
        for (int i = A.length; i >= 2; i--) {
            for (int j = 0; j < i - 1; j++) {
                if (A[j] == i) {
                    ans.add(j + 1);
                    flip(A, 0, j);
                    ans.add(i);
                    flip(A, 0, i - 1);
                    break;
                }
            }
        }
        return ans;
    }

    private void flip(int[] A, int i, int j) {
        while (i < j) {
            int tmp = A[i];
            A[i++] = A[j];
            A[j--] = tmp;
        }
    }
}