869.Reordered Power of 2

Description

Starting with a positive integer N, we reorder the digits in any order (including the original order) such that the leading digit is not zero.

Return true if and only if we can do this in a way such that the resulting number is a power of 2.

题目链接：https://leetcode.com/problems/reordered-power-of-2/

Difficulty: medium

Example 1:

Input: 1
Output: true

Example 2:

Input: 10
Output: false

Example 3:

Input: 16
Output: true

Example 4:

Input: 24
Output: false

Example 5:

Input: 46
Output: true

Note:

• 1 <= N <= 10^9

分析

1. updating

参考代码

class Solution:
def reorderedPowerOf2(self, N):
    li=[]
    for i in range(30):
        li.append(str(2**i))
    def judge(a,b):
        if(len(a) == len(b)):
            A=list(a)
            B=list(b)
            for index in A:
                if(index in B):
                    B.remove(index)
                else:
                    return False
            return True
        return False
    for i in range(len(li)):
        if(judge(li[i],str(N))):
            return True
    return False