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 <= 100A[i]is a permutation of[1, 2, ..., A.length]
Solution: Simulation
Put the largest element to its position. Each element requires two flips
e.g. [3, 2, 4, 1]
largest element: 4, index: 2
flip1: [4, 2, 3, 1]
flip2: [1, 3, 2, 4]
Repeat for [1, 3, 2]…
Time complexity: O(n^2)
Space complexity: O(1)
C++
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
// Author: Huahua, running time: 8 ms class Solution { public: vector<int> pancakeSort(vector<int>& A) { const int n = A.size(); vector<int> ans; for (int i = 0; i < n - 1; ++i) { int p = max_element(begin(A), end(A) - i) - begin(A); ans.push_back(p + 1); std::reverse(begin(A), begin(A) + p + 1); ans.push_back(n - i); std::reverse(begin(A), begin(A) + n - i); } return ans; } }; |
请尊重作者的劳动成果，转载请注明出处！花花保留对文章／视频的所有权利。
如果您喜欢这篇文章／视频，欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
huahualeetcode
Be First to Comment