969. Pancake Sorting

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 {
    public List<Integer> pancakeSort(int[] A) {
        List<Integer> res = new ArrayList();
        int max = A.length;
        for(int i = A.length - 1; i > -1; i--) {
            int ind = find(A, max);
            if(i == ind) {
                max--;
                continue;
            }
            else {
                if(ind != 0) res.add(ind + 1);//判断可有可无
                reverse(A, 0, ind);
                res.add(max);
                reverse(A, 0, --max);
            }
            
        }
        return res;
    }
    public int find(int[] A, int tar) {
        for(int i = 0; i < A.length; i++) {
            if(A[i] == tar) return i;
        }
        return -1;
    }
    public void reverse(int[] A, int l, int r) {
        while(l < r) {
            int tmp = A[l];
            A[l] = A[r];
            A[r] = tmp;
            l++;
            r--;
        }
    }
}

从大到小，找element，注意题目明确了数字范围是【1，length】，找到后先把ind + 1添加到res，然后反转【0，index】，然后添加element，反转【0，element-1】