算法

位运算

1. 2的幂

最右侧的1： `rightOne = m & (-m);

​```cpp
class Solution {
public:
    bool isPowerOfTwo(int n) {
        if (n == 0) return false;
        long m = (long)n;
        long rightOne = m & (-m);  //找到该数最右侧的1
        return rightOne == m ? true : false;
    }
};

2. 大于等于某个数的最大2的幂

class Solution {
public:
	bool NearTwoPower(int n) {
		if (n <= 0) return 1;
		n--; // 此处减一的目的是能够返回等于2的幂的数
		n |= n >> 1;
		n |= n >> 2;
		n |= n >> 4;
		n |= n >> 8;
		n |= n >> 16;
		return n + 1;
		
		//n = 00100000
		//    00011111 (--)
		//    00001111 (>>1)
		//n = 00011111
		//...
	}
}

3. 区间与的结果

class Solution {

public:
    int rangeBitwiseAnd(int left, int right) {
        while (left < right) {
            right -= right & -right; //每次去掉最右侧的1
        }
        return right;
    }
};

4. 数字逆序
   分治依次1/2/4/8交换

   class Solution {
   public:
    int ReverseNum (int n) {
     n = ((n & 0xAAAAAAAA) >> 1) | ((n & 0x55555555) << 1);
     n = ((n & 0xCCCCCCCC) >> 2) | ((n & 0x33333333) << 2);
     n = ((n & 0xF0F0F0F0) >> 4) | ((n & 0x0F0F0F0F) << 4);
     n = ((n & 0xFF00FF00) >> 8) | ((n & 0X00FF00FF) << 8);
     n = (n>> 16) | (n << 16);
     return n;
    }
   }