leetcode 1806.Minimum Number of Operations to Reinitialize a Permutation（python）

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描述

You are given an even integer n​​​​​​. You initially have a permutation perm of size n​​ where perm[i] == i​ (0-indexed)​​​​.

In one operation, you will create a new array arr, and for each i:

• If i % 2 == 0, then arr[i] = perm[i / 2].
• If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2].

You will then assign arr​​​​ to perm.

Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.

Example 1:

Input: n = 2
Output: 1
Explanation: perm = [0,1] initially.
After the 1st operation, perm = [0,1]
So it takes only 1 operation.

Example 2:

Input: n = 4
Output: 2
Explanation: perm = [0,1,2,3] initially.
After the 1st operation, perm = [0,2,1,3]
After the 2nd operation, perm = [0,1,2,3]
So it takes only 2 operations.

Example 3:

Input: n = 6
Output: 4

Note:

2 <= n <= 1000
n​​​​​​ is even

解析

根据题意，就是给出了一个偶数 n ，最初可以形成一个列表 perm ，以 0 为索引每个元素为 perm[i] == i​ , 我们进行一种特殊的操作，求至少做多少次这种操作，可以让 perm 又重新变回 perm 。这种操作就是：

• If i % 2 == 0, then arr[i] = perm[i / 2].
• If i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2]

按照题意，思路比较简单，就是直接按照这种操作，不断进行，并使用计数器 count 计数，当 perm 重新变回 perm 时，返回 count 即可。

解答

class Solution(object):
    def reinitializePermutation(self, n):
        """
        :type n: int
        :rtype: int
        """
        count = 0
        perm = [x for x in range(n)]
        result = [x for x in range(n)]
        while True:
            perm = [perm[i//2] if i % 2 ==0 else perm[n//2 + (i-1)//2] for i in range(n)]
            count += 1
            if perm==result:
                return count
        	      
		

运行结果

Runtime: 564 ms, faster than 38.89% of Python online submissions for Minimum Number of Operations to Reinitialize a Permutation.
Memory Usage: 13.3 MB, less than 77.78% of Python online submissions for Minimum Number of Operations to Reinitialize a Permutation.

解析

另外，可以使用递归来解决 perm 的不断变化。

解答

class Solution(object):
    def __init__(self):
        self.count = 0
    def reinitializePermutation(self, n):
        """
        :type n: int
        :rtype: int
        """
        perm = [x for x in range(n)]
        result = [x for x in range(n)]
        def f(perm):
            perm = [perm[i//2] if i % 2 ==0 else perm[n//2 + (i-1)//2] for i in range(n)]
            self.count += 1
            if result == perm:
                return
            f(perm)
        f(perm)
        return self.count

运行结果

Runtime: 600 ms, faster than 33.33% of Python online submissions for Minimum Number of Operations to Reinitialize a Permutation.
Memory Usage: 21.4 MB, less than 11.11% of Python online submissions for Minimum Number of Operations to Reinitialize a Permutation.

原题链接：leetcode.com/problems/mi…

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