231. Power of Two

Given an integer, write a function to determine if it is a power of two.

解法一：Language - Java Run Time - 2ms
解題思路：一個整數如果可以表示爲2的冪次，就可以表示爲100000……；從而這個數減去1就變爲011111……

 public boolean isPowerOfTwo(int n) {
        if(n > 0 && (n & (n-1)) == 0) { 
            return true;
        }else{
            return false;
        }
    }

或

 public boolean isPowerOfTwo(int n) {
        return n > 0 && (n & (n-1)) == 0;
    }

解法二：Language - Java Run Time - 2ms
解題思路：一個整數如果可以表示爲2的冪次，這個數的二進制表示中1的個數只有一個

 public boolean isPowerOfTwo(int n) {
        //n中1的個數爲1個
        return n > 0 && Integer.bitCount(n) == 1; 
    }

解法三：Language - Java Run Time - 2ms
解題思路：暴力枚舉

public boolean isPowerOfTwo(int n) {
        return n == 1 || n == 2 || n == 4 || n == 8 || n == 16 || n == 32 || n == 64 || n == 128 || n == 256 || n == 512 || n == 1024 || n == 2048 || n == 4096 || n == 8192 || n == 16384 || n ==32768 || n == 65536 || n == 131072 || n == 262144 || n == 524288 || n == 1048576 || n == 2097152 || n == 4194304 || n == 8388608 || n == 16777216 || n == 33554432 || n == 67108864 || n == 134217728 || n == 268435456 || n == 536870912 || n == 1073741824;
    }

解法四：Language - Java Run Time - 2ms
解題思路：遞歸

public boolean isPowerOfTwo(int n) {
        if(n <= 0) return false;
        if(n == 1) return true;
        //n爲奇數
        if(n % 2 != 0) return false;
        //遞歸調用
        return isPowerOfTwo(n/2);
    }

解法五：Language - Java Run Time - 2ms
解題思路：循環

public boolean isPowerOfTwo(int n) {
        if(n > 0){
            //n是2倍數
            while(n % 2 == 0){
                n /= 2;
            }
        }
        return n == 1;
    }