969. Pancake Sorting

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 kelements 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.lengthflips 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]

思路：从N遍历到1，每次先解决剩下的最大的数，比如遍历到10（10在第5位上），也就是说10-N已经在正确的位置上，1-10还没有。那就先flip 1-5把10放到第一位，接着flip 1-10把10放到正确的位置上。

保证2N次交换一定可以

class Solution:
    def pancakeSort(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        if tuple(A)==tuple(sorted(A)): return []
        d={v:i+1 for i,v in enumerate(A)}
        res = []
        for i in range(len(A),0,-1):
            res.append(d[i])
            res.append(i)
            
            d2={}
            for s in d:
                if s>i: continue
                pos=d[s]
                if pos<=d[i]:
                    pos=d[i]+1-pos
                pos=i+1-pos
                d2[s] = pos
            d=d2
        
        return res

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