面试题:以最快的速度计算一个字节（byte）里有多少bit被置1

本文详细分析并对比了四种计算整数中1的位数的算法，包括循环遍历、位操作优化、位掩码组合以及位运算快速计数方法，通过实例展示算法的原理和性能特点。

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算法一

int count_ones(unsigned a) { int count = 0; for (; a; a >>= 1) { if (a & 1) count ++; } return count; }

算法二

int count_ones(unsigned a) { int count = 0; for (; a; count ++) { a &= a - 1; } return count; }

算法三

int count_ones(unsigned a) { a = (a & 0x55555555) + ((a >> 1) & 0x55555555); a = (a & 0x33333333) + ((a >> 2) & 0x33333333); a = (a & 0x0f0f0f0f) + ((a >> 4) & 0x0f0f0f0f); a = (a & 0x00ff00ff) + ((a >> 8) & 0x00ff00ff); a = (a & 0x0000ffff) + ((a >> 16) & 0x0000ffff); return a; }

算法四

const int ones[] = {0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2,\ 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4,\ 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2,\ 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5,\ 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5,\ 6, 4, 5, 5, 6, 5, 6, 6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,\ 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3,\ 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 2, 3,\ 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5,\ 6, 5, 6, 6, 7, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 4, 5, 5, 6,\ 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8}; int count_ones(unsigned a) { return (ones[a & 0xff] + ones[(a >> 8) & 0xff] + ones[(a >> 16) & 0xff]\ + ones[(a >> 24) & 0xff]); }