1. 如果n为2的幂，则
a%n = a & (n-1)
2. mod运算规则：
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结合律
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((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
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交换律
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(a + b) mod p = (b+a) mod p
(a × b) mod p = (b × a) mod p
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分配律
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((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
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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); }