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 }总结
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