这是确实可能的,只需使用
原地基数排序即可。
它运行时间为
O(kn)
,其中
k
对于任何标准数字数据类型都是常数,并且需要
O(1)
额外空间。
以下是代码:
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
void I32R(int32_t *data, uint32_t size, uint32_t nbit) {
uint32_t dbgn = (uint32_t)-1, dend = size;
while (++dbgn < dend)
if (data[dbgn] & nbit)
while (dbgn < --dend)
if (~data[dend] & nbit) {
data[dbgn] ^= data[dend];
data[dend] ^= data[dbgn];
data[dbgn] ^= data[dend];
break;
}
if ((nbit >>= 1) && (dend > 1))
I32R(data, dend, nbit);
if (nbit && (size - dend > 1))
I32R(data + dend, size - dend, nbit);
}
int32_t dups(int32_t *data, uint32_t size) {
int32_t iter, *uniq = data;
if (size < 2)
return size;
for (iter = 0; iter < size; iter++)
data[iter] ^= (1 << 31);
I32R(data, size, 1 << 31);
data[0] ^= (1 << 31);
for (iter = 1; iter < size; iter++)
if (*uniq != (data[iter] ^= (1 << 31)))
*++uniq = data[iter];
return uniq - data + 1;
}
void parr(int32_t *data, uint32_t size) {
for (; size; size--)
printf("%4d%s", *data++, (size == 1)? "\n\n" : ", ");
}
int main() {
int32_t iter, size, *data;
data = malloc((size = 256) * sizeof(*data));
for (iter = 0; iter < size; iter++)
data[iter] = (int8_t)rand() & -3;
parr(data, size);
parr(data, dups(data, size));
free(data);
return 0;
}
注意1:在排序之前反转符号位对于使正数大于负数是必要的,因为基数排序仅对无符号值进行操作。
注意2:这只是一个粗略的示例,从未真正测试过。
注意3:哇,这实际上比qsort()
更快!
注意4:现在有一个非递归版本的排序函数;用法几乎相同,除了没有nbit
外:
void I32NR(int32_t *data, uint32_t size) {
int32_t mask, head;
struct {
uint32_t init, size, nbit, edge;
} heap[32];
heap[0].nbit = 32;
heap[0].size = size;
heap[0].init = head = 0;
do {
size = heap[head].init - 1;
mask = 1 << ((heap[head].nbit & 0x7F) - 1);
heap[head].edge = heap[head].size;
while (++size < heap[head].edge)
if (data[size] & mask)
while (size < --heap[head].edge)
if (~data[heap[head].edge] & mask) {
data[size] ^= data[heap[head].edge];
data[heap[head].edge] ^= data[size];
data[size] ^= data[heap[head].edge];
break;
}
heap[head].nbit = ((heap[head].nbit & 0x7F) - 1)
| (heap[head].nbit & 0x80);
if ((heap[head].nbit & 0x7F) && (heap[head].edge > 1)) {
heap[head + 1] = heap[head];
heap[head + 1].size = heap[head].edge;
heap[++head].nbit |= 0x80;
continue;
}
do {
if ((heap[head].nbit & 0x7F)
&& (heap[head].size - heap[head].edge > 1)) {
heap[head + 1] = heap[head];
heap[head + 1].init = heap[head].edge;
heap[++head].nbit &= 0x7F;
break;
}
while ((head >= 0) && !(heap[head--].nbit & 0x80));
} while (head >= 0);
} while (head >= 0);
}
bool
数组,但那会占用空间。我真的需要它完成。 - frederick99