位操作

1. 如果n为2的幂，则

a%n = a & (n-1)

2. mod运算规则：

结合律	((a+b) mod p + c)mod p = (a + (b+c) mod p) mod p ((a*b) mod p * c)mod p = (a * (b*c) mod p) mod p
交换律	(a + b) mod p = (b+a) mod p (a × b) mod p = (b × a) mod p
分配律	((a +b)mod p × c) mod p = ((a × c) mod p + (b × c) mod p) mod p (a×b) mod c=(a mod c * b mod c) mod c (a+b) mod c=(a mod c+ b mod c) mod c (a-b) mod c=(a mod c- b mod c) mod c

3. 判断一个数n是否是2的幂：

n > 0 && ((n & (n - 1)) == 0 )

4. 最高位1对应的整数：

    /**
     * Returns an {@code int} value with at most a single one-bit, in the
     * position of the highest-order ("leftmost") one-bit in the specified
     * {@code int} value.  Returns zero if the specified value has no
     * one-bits in its two's complement binary representation, that is, if it
     * is equal to zero.
     *
     * @return an {@code int} value with a single one-bit, in the position
     *     of the highest-order one-bit in the specified value, or zero if
     *     the specified value is itself equal to zero.
     * @since 1.5
     */
    public static int highestOneBit(int i) {
        // HD, Figure 3-1
        i |= (i >>  1);
        i |= (i >>  2);
        i |= (i >>  4);
        i |= (i >>  8);
        i |= (i >> 16);
        return i - (i >>> 1);
    }

5. 最低位1对应的整数：

    /**
     * Returns an {@code int} value with at most a single one-bit, in the
     * position of the lowest-order ("rightmost") one-bit in the specified
     * {@code int} value.  Returns zero if the specified value has no
     * one-bits in its two's complement binary representation, that is, if it
     * is equal to zero.
     *
     * @return an {@code int} value with a single one-bit, in the position
     *     of the lowest-order one-bit in the specified value, or zero if
     *     the specified value is itself equal to zero.
     * @since 1.5
     */
    public static int lowestOneBit(int i) {
        // HD, Section 2-1
        return i & -i;
    }

其实是i&(i-1的补数)，因为i的补数加1等于-i，所以i的补数=-(i+1)，i-1的补数=-i。

i+i的补数=0xFFFFFFFF，i+(-i-1)=0xFFFFFFFF

6. 前面0的个数：

    public static int numberOfLeadingZeros(int i) {
        // HD, Figure 5-6
        if (i == 0)
            return 32;
        int n = 1;
        if (i >>> 16 == 0) { n += 16; i <<= 16; }
        if (i >>> 24 == 0) { n +=  8; i <<=  8; }
        if (i >>> 28 == 0) { n +=  4; i <<=  4; }
        if (i >>> 30 == 0) { n +=  2; i <<=  2; }
        n -= i >>> 31;
        return n;
    }

显然采用了二分的算法，不过从计算机的角度，移两次位应该是移一次位的两倍时间（猜测），这样的二分真的合适吗

7. 后面0的个数：

    public static int numberOfTrailingZeros(int i) {
        // HD, Figure 5-14
        int y;
        if (i == 0) return 32;
        int n = 31;
        y = i <<16; if (y != 0) { n = n -16; i = y; }
        y = i << 8; if (y != 0) { n = n - 8; i = y; }
        y = i << 4; if (y != 0) { n = n - 4; i = y; }
        y = i << 2; if (y != 0) { n = n - 2; i = y; }
        return n - ((i << 1) >>> 31);
    }