[Swift Weekly Contest 118]LeetCode969.煎饼排序 | Pancake Sor

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: [4,3] Explanation: We perform 4 pancake flips,with k values 4,and 3. Starting state: A = [3,1] After 1st flip (k=4): A = [1,3] After 2nd flip (k=2): A = [4,1,3] After 3rd flip (k=4): A = [3,4] After 4th flip (k=3): A = [1,3,4],which is sorted. 

Input: [1,3]
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. 1 <= A.length <= 100
2. A[i]?is a permutation of?[1,...,A.length]

给定数组?A，我们可以对其进行煎饼翻转：我们选择一些正整数?k?<= A.length，然后反转?A?的前?k?个元素的顺序。我们要执行零次或多次煎饼翻转（按顺序一次接一次地进行）以完成对数组?A?的排序。

返回能使?A?排序的煎饼翻转操作所对应的 k 值序列。任何将数组排序且翻转次数在?10 * A.length?范围内的有效答案都将被判断为正确。

示例 1：

输入：[3,1]
输出：[4,3]
解释：
我们执行 4 次煎饼翻转，k 值分别为 4，2，4，和 3。
初始状态 A = [3,1]
第一次翻转后 (k=4): A = [1,3]
第二次翻转后 (k=2): A = [4,3]
第三次翻转后 (k=4): A = [3,4]
第四次翻转后 (k=3): A = [1,4]，此时已完成排序。 

示例 2：

输入：[1,3]
输出：[]
解释：
输入已经排序，因此不需要翻转任何内容。
请注意，其他可能的答案，如[3，3]，也将被接受。

提示：

1. 1 <= A.length <= 100
2. A[i]?是?[1,A.length]?的排列

40ms

 1 class Solution {
 2     func pancakeSort(_ A: [Int]) -> [Int] {
 3         var A = A
 4         var n:Int = A.count
 5         var ans:[Int] = [Int]()
 6         for i in (0...(n - 1)).reversed()
 7         {
 8             var j:Int = 0
 9             while(A[j] != i+1)
10             {
11                 j += 1
12             }
13             ans.append(j + 1)
14             ans.append(i + 1)
15             var newA:[Int] = [Int](repeating:0,count:n)
16             for k in 0..<n
17             {
18                 newA[k] = A[k]
19             }
20             for k in 0...j
21             {
22                 newA[k] = A[j-k]
23             }
24             A = newA
25             newA = [Int](repeating:0,count:n)
26             for k in 0..<n
27             {
28                 newA[k] = A[k]
29             }
30             for k in 0...i
31             {
32                 newA[k] = A[i-k]
33             }
34             A = newA
35         }
36         return ans
37     }
38 }