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c - 如何合并不同大小的已排序序列?

转载 作者:行者123 更新时间:2023-11-30 18:55:04 25 4
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我正在寻找一种算法来将多个排序序列(比如说 X 个排序序列)合并到 c 中的一个排序序列中,您能提供一些示例吗?

请注意,每个序列都有不同的大小。

我想到了这一点,但序列可能有重叠的间隔。所以它不会工作。

x=s1[first] 
y=s2[last]
if(x<y) then put s1 after s2 in the output.
...

注意:我不想使用任何库

最佳答案

您可以像 N 路 merge sort 一样合并输入序列确实如此。

基本上,您检查每个有序输入序列的头部,并将最小头部附加到新的有序序列中(并更新您使用的序列的头部)。这样做直到所有输入序列都为空。

如果您有大量的输入序列需要合并,那么您可能需要维护输入序列头部的最小堆,而不是每次都对最小头部进行线性扫描。

#include <stddef.h>
#include <stdlib.h>
#include <stdio.h>
#include <assert.h>

typedef struct
{
size_t size;
int *seq;

} Sequence;

Sequence *mergeSortedSequences(const Sequence *seqs[], size_t num_seqs)
{
Sequence *ret = malloc(sizeof(Sequence));
size_t indexes[num_seqs]; /* NOTE: not valid C89; in C89 need to malloc instead or have seqs be non-const and trash the Sequences themselves */
size_t tot_size;
size_t i;

if (NULL == ret)
goto FAIL;

ret->size = 0;

/* calculate total size of merged Sequence, init indexes and cull any empty sequences from seqs */

for (i = 0, tot_size = 0; i < num_seqs;)
{
if (0 != seqs[i]->size)
{
tot_size += seqs[i]->size;
indexes[i] = 0;
++i;
}
else
{
seqs[i] = seqs[--num_seqs]; /* cull empty seq from seqs */
printf("Swapping original seq %lu into position %lu\n", num_seqs, i);
}
}

printf("Return sequence size is %lu; number of non-empty sequences: %lu\n", tot_size, num_seqs);

ret->seq = calloc(tot_size + 1, sizeof(int)); /* NOTE: +1 so we don't ever calloc(0) */

if (NULL == ret->seq)
goto FAIL_RET;

/* loop while non-empty sequences left */

while (num_seqs)
{
/* find minimum of remaining seqs */

int min = seqs[0]->seq[indexes[0]];
size_t min_seq = 0;

for (i = 1; i < num_seqs; ++i)
{
if (seqs[i]->seq[indexes[i]] < min)
{
min = seqs[i]->seq[indexes[i]];
min_seq = i;
}
}

/* update ret */

ret->seq[ret->size++] = min;

printf("minimum value was from head of seq %lu; ret[%lu] = %d\n", min_seq, ret->size, min);

/* update indexes, min_seq and seqs */

++indexes[min_seq]; /* "consume" the head of min_seq */

if (indexes[min_seq] == seqs[min_seq]->size) /* we've exhausted min_seq */
{
seqs[min_seq] = seqs[--num_seqs]; /* cull empty sequence */
indexes[min_seq] = indexes[num_seqs];
printf("Swapping seq %lu into position %lu (2)\n", num_seqs, min_seq);
}
}

printf("Success ret->size = %lu\n", ret->size);

assert(NULL != ret && ret->size == tot_size);
goto END;

/* error handling and return */

FAIL_RET:
free(ret);
ret = NULL;

FAIL:
assert(NULL == ret);

END:
return ret;
}

int main(int argc, char **argv)
{
int arr1[] = { -44, 0, 22, 49, 58, 101 };
int arr3[] = { -101, 14, 22, 31, 151 };
int arr4[] = { -1, 0, 1, 50, 200 };
Sequence seq0 = { 0, NULL };
Sequence seq1 = { sizeof(arr1) / sizeof(arr1[0]) };
Sequence seq2 = { 0, NULL };
Sequence seq3 = { sizeof(arr3) / sizeof(arr3[0]) };
Sequence seq4 = { sizeof(arr4) / sizeof(arr4[0]) };
Sequence seq5 = { 0, NULL };
Sequence *seqs[6];
Sequence *ret;
size_t i;

seq1.seq = arr1;
seq3.seq = arr3;
seq4.seq = arr4;

seqs[0] = &seq0;
seqs[1] = &seq1;
seqs[2] = &seq2;
seqs[3] = &seq3;
seqs[4] = &seq4;
seqs[5] = &seq5;

ret = mergeSortedSequences((const Sequence**) seqs, sizeof(seqs) / sizeof(seqs[0]));

if (NULL == ret)
{
perror("Allocation failure!");
abort();
}

printf("Merged sequence contains %lu elements:\n", ret->size);

for (i = 0; i < ret->size; ++i)
printf("\t%d\n", ret->seq[i]);

return 0;
}

关于c - 如何合并不同大小的已排序序列?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28648633/

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