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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]
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:[3,2,4,1] 输出:[4,2,4,3] 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, 2, 4, 1] 第一次翻转后 (k=4): A = [1, 4, 2, 3] 第二次翻转后 (k=2): A = [4, 1, 2, 3] 第三次翻转后 (k=4): A = [3, 2, 1, 4] 第四次翻转后 (k=3): A = [1, 2, 3, 4],此时已完成排序。
示例 2:
输入:[1,2,3] 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, 2, ..., A.length]
的排列
16ms
1 class Solution { 2 func pancakeSort(_ A: [Int]) -> [Int] { 3 var aCopy = A 4 var result = [Int]() 5 var valueToBeSort = A.count 6 7 while valueToBeSort != 1 { 8 if valueToBeSort == aCopy[valueToBeSort - 1] { 9 valueToBeSort -= 1 10 } else { 11 let index = aCopy.firstIndex(of: valueToBeSort)! 12 if index != 0 { 13 result.append(index + 1) 14 for i in 0...index / 2 { 15 aCopy.swapAt(i, index - i) 16 } 17 } 18 result.append(valueToBeSort) 19 for i in 0...(valueToBeSort - 1) / 2 { 20 aCopy.swapAt(i, valueToBeSort - 1 - i) 21 } 22 } 23 } 24 25 return result 26 } 27 }
24ms
1 class Solution { 2 func pancakeSort(_ A: [Int]) -> [Int] { 3 var A = A 4 var ans = [Int]() 5 var i = A.count 6 while i>0 { 7 var j=0 8 for k in 0..<A.count { 9 if A[k] == i { 10 j = k 11 break 12 } 13 } 14 flip(&A, &ans, j); 15 flip(&A, &ans, i-1); 16 i -= 1 17 } 18 return ans 19 } 20 21 func flip(_ A: inout [Int], _ ans: inout [Int], _ pos: Int) -> Void { 22 var s = 0, e = pos 23 while s<e { 24 let temp=A[s] 25 A[s]=A[e] 26 A[e]=temp 27 s += 1 28 e -= 1 29 } 30 ans.append(pos+1) 31 } 32 }
32ms
1 class Solution { 2 func pancakeSort(_ A: [Int]) -> [Int] { 3 var B = A 4 var n = A.count 5 var ret = [Int]() 6 while n >= 1 { 7 let i = B.firstIndex(of: n)! 8 B[0...i].reverse() 9 ret.append(i+1) 10 B[0..<n].reverse() 11 ret.append(n) 12 n -= 1 13 } 14 return ret 15 } 16 }
36ms
1 class Solution { 2 func pancakeSort(_ A: [Int]) -> [Int] { 3 let sorted = A.sorted() 4 guard A != sorted else { return [] } 5 var ar = A 6 var lastSorted = sorted.count - 1 7 var res: [Int] = [] 8 while lastSorted >= 0 { 9 let max = sorted[lastSorted] 10 let ind = ar.index(of: max)! 11 if ind == lastSorted { 12 } else { 13 res.append(ind + 1) 14 res.append(lastSorted + 1) 15 var ar1 = ar 16 if ind != 0 { 17 ar1 = Array(Array(ar[0...ind]).reversed()) + Array(ar[ind+1..<ar.count]) 18 } 19 if lastSorted == ar1.count - 1 { 20 ar = Array(ar1.reversed()) 21 } else { 22 ar = Array(Array(ar1[0...lastSorted]).reversed()) + Array(ar1[lastSorted+1..<ar1.count]) 23 } 24 } 25 lastSorted -= 1 26 } 27 28 return res 29 } 30 }
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 }
80ms
1 class Solution { 2 func pancakeSort(_ A: [Int]) -> [Int] { 3 4 var res: [Int] = [] 5 6 var sortArray: [Int] = A.sorted() 7 var A = A 8 9 for i in (0...A.count - 1).reversed() { 10 if A[i] == sortArray[i] { 11 continue 12 } else { 13 for j in (0...i).reversed() { 14 if A[j] == sortArray[i] { 15 A.replaceSubrange(Range.init(0...j), with: A[0...j].reversed()) 16 A.replaceSubrange(Range.init(0...i), with: A[0...i].reversed()) 17 18 res.append(j + 1) 19 res.append(i + 1) 20 21 break 22 } 23 } 24 } 25 26 } 27 28 return res 29 } 30 }
100ms
1 class Solution { 2 func pancakeSort(_ A: [Int]) -> [Int] { 3 // find largest and send to back 4 var sorted = true 5 for i in 0..<A.count - 1 { 6 if A[i] > A[i + 1] { 7 sorted = false 8 break; 9 } 10 } 11 if sorted { return [] } 12 13 var flippedArr = A 14 var accum: [Int] = [] 15 for i in 0..<A.count { 16 var max = biggest(A.count - 1 - i, flippedArr) 17 accum.append(max + 1) 18 flippedArr = flip(max, flippedArr) 19 print(flippedArr) 20 accum.append(A.count - i) 21 flippedArr = flip(A.count - 1 - i, flippedArr) 22 // print(flippedArr) 23 } 24 return accum 25 } 26 27 func flip(_ end: Int, _ arr: [Int]) -> [Int] { 28 var out = arr 29 for i in 0...end/2 { 30 let temp = out[i] 31 out[i] = out[end - i] 32 out[end - i] = temp 33 } 34 35 return out 36 } 37 38 func biggest(_ end: Int, _ arr: [Int]) -> Int { 39 var max = 0 40 for i in 0...end { 41 if arr[i] >= arr[max] { 42 max = i 43 } 44 } 45 return max 46 } 47 }