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

概述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 (doi

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. 

Example 2:

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 <= A.length <= 100 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 *** 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 <= A.length <= 100 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 += 112             }13             ans.append(j + 1)14             ans.append(i + 1)15             var newA:[Int] = [Int](repeating:0,count:n)16             for k in 0..<n17             {18                 newA[k] = A[k]19             }20             for k in 0...j21             {22                 newA[k] = A[j-k]23             }24             A = newA25             newA = [Int](repeating:0,count:n)26             for k in 0..<n27             {28                 newA[k] = A[k]29             }30             for k in 0...i31             {32                 newA[k] = A[i-k]33             }34             A = newA35         }36         return ans37     }38 }

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