标签：col == bre tput any lis pre osi fast
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 <= N <= 10^9
class Solution { private char[] Ncarr; public boolean ispermutation(String astr){ char[] acarr = astr.toCharArray(); Arrays.sort(acarr); for(int i=0; i<acarr.length; i++){ if(acarr[i] != Ncarr[i]) return false; } return true; } public boolean reorderedPowerOf2(int N) { if(N == 1 || N == 2) return true; String Nstr = Integer.toString(N); Ncarr = Nstr.toCharArray(); Arrays.sort(Ncarr); int digitN = Nstr.length(); int base = 2; List<String> tmplist = new ArrayList<>(); while(true){ String tmp = Integer.toString(base); if(tmp.length() == digitN) tmplist.add(tmp); if(((base >> 31) & 1) == 1) break; if(tmp.length() > digitN) break; base <<= 1; } for(String x : tmplist){ if(ispermutation(x)) return true; } return false; } }
标签：col == bre tput any lis pre osi fast
原文地址：https://www.cnblogs.com/ethanhong/p/10262413.html
