拆分数组。 [英] splitting an array.
问题描述
我有一个数组：
{100,20，-45 -345，-2 120,64,99,20,15,0,1,25}
我想把它分成两个不同的数组，这样每个数字< =
50进入左数组，每个数字50进入右数组。
我做了一些编码，但我觉得这段代码效率很低：
void split_array（int * a，int size_of_array）
{
/ * a是指向要分区的数组的指针* /
int i，left_size = 0，right_size = 0;
int * b，* c / *指向新数组的指针* /
for（i = 0; i< size_of_array; i ++）
{
if（a [i]< = 50）
left_size ++;
if（a [i ] 50）
right_size ++;
}
b = calloc（sizeof（* b）* left_size）;
c = calloc（sizeof（* c）* right_size）;
if（b == NULL || c == NULL）
{
fprintf（stderr，内存分配失败：％s％d％s"，__ FILE __，
__LINE __，__ func__）;
退出（EXIT_FAILURE）;
}
left_size = right_size = 0;
for（i = 0; I< size_of_array; i ++）
{
if（a [i]< = 50）
{
b [left_size] = a [i];
left_size ++;
}
if（a [i] 50）
{
c [right_size] = a [i];
right_size ++;
}
}
退出（EXIT_SUCCESS）;
}
我真的不喜欢运行类似的for循环二时间。
这是不好的编程吗？
I''ve an array :
{100,20, -45 -345, -2 120, 64, 99, 20, 15, 0, 1, 25}
I want to split it into two different arrays such that every number <=
50 goes into left array and every number 50 goes into right array.
I''ve done some coding but I feel this code is very inefficient:
void split_array(int *a, int size_of_array)
{
/* a is the pointer to the array which is going to be partitioned */
int i, left_size =0, right_size = 0;
int *b, *c /* pointers to new arrays */
for(i =0; i< size_of_array; i++)
{
if(a[i] <= 50)
left_size++;
if(a[i] 50)
right_size++;
}
b = calloc(sizeof(*b) * left_size);
c = calloc(sizeof(*c) * right_size);
if( b == NULL || c == NULL)
{
fprintf(stderr, "memory allocation failure: %s %d %s", __FILE__,
__LINE__, __func__);
exit(EXIT_FAILURE);
}
left_size = right_size = 0;
for(i =0; i< size_of_array; i++)
{
if(a[i] <= 50)
{
b[left_size] = a[i];
left_size++;
}
if(a[i] 50)
{
c[right_size] = a[i];
right_size++;
}
}
exit(EXIT_SUCCESS);
}
I''m really not comfortable with running similar for loops two times.
Is this bad programming ?
推荐答案
pereges写道：
pereges wrote:
我有一个数组：
{100,20，-45 -345，-2 120,64,99,20,15,0,1， 25}
我想把它分成两个不同的数组，每个数字< =
50进入左数组，每个数字50进入右边阵列。
I''ve an array :
{100,20, -45 -345, -2 120, 64, 99, 20, 15, 0, 1, 25}
I want to split it into two different arrays such that every number <=
50 goes into left array and every number 50 goes into right array.
/ * BEGIN new.c输出* /
原始数组：
100 20 -45 -345 -2 120 64 99 20 15 0 1 25
剩余阵列：
-345 -45 -2 0 1 15 20 20 25
右阵：
>
/ * BEGIN new.c * /
#include< stdio.h>
#include< stdlib .h>
#define权利50
int compar（const void *，const void *）;
int main（无效）
{
size_t count;
int array [] = {100,20，-45 ，-345，-2,120,64,99,20,15,0,1,25};
int * right = array;
puts（ " / * BEGIN new.c output * / \ n"）;
puts（" original array："）;
for（count = 0; count！= sizeof array / sizeof * array; ++ count）{
printf（"％d"，array [count]）;
}
putchar（''\ n''）;
qs ort（array，sizeof array / sizeof * array，sizeof * array，compar）;
while（RIGHT * right
&& right！= array + sizeof array / sizeof * array）
{
++ right;
}
puts（" \ nleft array："）;
for（count = 0; count！= right - array + 0u; ++ count）{
printf （"％d"，array [count]）;
}
puts（" \ n\\\
right array："）;
for（count = 0;
count！= sizeof array / sizeof * array - （right - array）; ++ count）
{
printf（"％d"，right [count]）;
}
puts（" \ n\\\
/ * END new.c输出* /"）;
返回0;
}
int compar（const void * a ，const void * b）
{
const int * pa = a;
const int * pb = b;
返回* pb * pa？ -1：* pb！= * pa;
}
/ * END new.c * /
-
pete
/* BEGIN new.c output */
original array:
100 20 -45 -345 -2 120 64 99 20 15 0 1 25
left array:
-345 -45 -2 0 1 15 20 20 25
right array:
64 99 100 120
/* END new.c output */
/* BEGIN new.c */
#include <stdio.h>
#include <stdlib.h>
#define RIGHT 50
int compar(const void *, const void *);
int main(void)
{
size_t count;
int array[] = {100,20,-45,-345,-2,120,64,99,20,15,0,1,25};
int *right = array;
puts("/* BEGIN new.c output */\n");
puts("original array:");
for (count = 0; count != sizeof array / sizeof *array; ++count) {
printf("%d ", array[count]);
}
putchar(''\n'');
qsort(array, sizeof array / sizeof *array, sizeof *array, compar);
while (RIGHT *right
&& right != array + sizeof array / sizeof *array)
{
++right;
}
puts("\nleft array:");
for (count = 0; count != right - array + 0u; ++count) {
printf("%d ", array[count]);
}
puts("\n\nright array:");
for (count = 0;
count != sizeof array / sizeof *array - (right - array); ++count)
{
printf("%d ", right[count]);
}
puts("\n\n/* END new.c output */");
return 0;
}
int compar (const void *a, const void *b)
{
const int *pa = a;
const int *pb = b;
return *pb *pa ? -1 : *pb != *pa;
}
/* END new.c */
--
pete
Keith Thompson说：
Keith Thompson said:
pereges< Br *** **@gmail.comwrites：
pereges <Br*****@gmail.comwrites:
< snip>
<snip>
> if（a [i]< = 50）
left_size ++;
if（a [i] 50）
right_size ++;
> if(a[i] <= 50)
left_size++;
if(a[i] 50)
right_size++;
第二次测试是不必要的。
The second test is unnecessary.
s /是不必要的/可以替换为else /
< snip>
-
Richard Heathfield< http://www.cpax.org.uk>
电子邮件：-http：// www。 + rjh @
谷歌用户：< http://www.cpax.org.uk/prg/writings/googly.php>
Usenet是一个奇怪的放置" - dmr 1999年7月29日
s/is unnecessary/can be replaced by else/
<snip>
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
Richard Heathfield< rj*@see.sig.invalidwrites：
Richard Heathfield <rj*@see.sig.invalidwrites:
Keith汤普森说：
Keith Thompson said:
> pereges< Br ***** @ gmail.comwrites：
>pereges <Br*****@gmail.comwrites:
< snip>
<snip>
>> if（a [i]< = 50）
left_size ++;
if（a [i] 50）
right_size ++;
>> if(a[i] <= 50)
left_size++;
if(a[i] 50)
right_size++;
第二次测试是不必要的。
The second test is unnecessary.
s /是不必要的/可以用else替换/
< snip>
s/is unnecessary/can be replaced by else/
<snip>
正确。
-
Keith Thompson（The_Other_Keith） ks *** @ mib.org < http://www.ghoti.net/~kst>
Nokia
我们必须做点什么。这是事情。因此，我们必须这样做。
- Antony Jay和Jonathan Lynn，是部长
Right.
--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
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