leetcode brush 9 title

Problems encountered today is LeetCode 231 questions. The problem is a simple question, asked to enter a number n, determine the number is not a power of 2
to see this problem, the first thought is to use bit operation, because binary bit computing is more efficient
my own solution It is the following code:

public static boolean judge(int n){
    int num=1;
    boolean flage=false;
    while (num<=n){
        if (num==n){
            flage=true;
           break;
        }else {
            num=num<<1;
        }
    }
    return flage;
}            

However, this scheme n particularly when time out. After the reference to other people's answers to the program, I found that if a number n is a power of 2, then the number must meet three conditions:

① the highest binary number is 1, the other bits are 0;
② the number minus one of the most significant bit is 0, the other bit is a 1;
③ The number must be greater than 0
and therefore can be utilized in the position calculation, as follows:

public static boolean judge2(int n){
    //if (n<=0) return false;
    //return ((n&(n-1))==0);
    //或者写成
    return n>0 && (n&(n-1))==0;
}