我正在寻找比下面更快的算法。给定一个 64 位无符号整数序列，返回该序列中每个 64 位被设置的次数计数。
Example:
4608 = 0000000000000000000000000000000000000000000000000001001000000000
4097 = 0000000000000000000000000000000000000000000000000001000000000001
2048 = 0000000000000000000000000000000000000000000000000000100000000000
counts 0000000000000000000000000000000000000000000000000002101000000001
Example:
2560 = 0000000000000000000000000000000000000000000000000000101000000000
530 = 0000000000000000000000000000000000000000000000000000001000010010
512 = 0000000000000000000000000000000000000000000000000000001000000000
counts 0000000000000000000000000000000000000000000000000000103000010010
目前我正在使用一种相当明显且幼稚的方法：
static int bits = sizeof(ulong) * 8;
public static int[] CommonBits(params ulong[] values) {
int[] counts = new int[bits];
int length = values.Length;
for (int i = 0; i < length; i++) {
ulong value = values[i];
for (int j = 0; j < bits && value != 0; j++, value = value >> 1) {
counts[j] += (int)(value & 1UL);
}
}
return counts;
}
通过首先将整数进行或运算，然后使用结果来确定需要检查哪些位，可以实现较小的速度改进。您仍然需要迭代每个位，但只迭代一次没有 1 的位，而不是values.Length times.
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