969

969. Pancake Sorting (Medium)

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

Companies:
Square

Related Topics:
Array, Sort

Solution 1.

// OJ: https://leetcode.com/problems/pancake-sorting
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    vector<int> pancakeSort(vector<int>& A) {
        vector<int> ans;
        for (int i = A.size(); i > 0; --i) {
            int j = i - 1;
            for (; j >= 0 && A[j] != i; --j);
            reverse(A.begin(), A.begin() + j + 1);
            ans.push_back(j + 1);
            reverse(A.begin(), A.begin() + i);
            ans.push_back(i);
        }
        return ans;
    }
};