869. Reordered Power of 2

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.

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. 1 <= N <= 10^9

Approach #1: Math. [Java]

class Solution {
    public boolean reorderedPowerOf2(int N) {
        int c = count(N);
        for (int i = 0; i < 32; ++i) {
            if (count(1 << i) == c) return true;
        }
        return false;
    }
    
    public int count(int x) {
        int ret = 0;
        for (; x > 0; x /= 10)
            ret += (int)Math.pow(10, x % 10);
        return ret;
    }
}

Analysis:

The way that use / and % to count the digit is awesome.

Reference:

https://leetcode.com/problems/reordered-power-of-2/discuss/149843/C%2B%2BJavaPython-Straight-Forward

869. Reordered Power of 2

原文：https://www.cnblogs.com/ruruozhenhao/p/10897015.html