拆分数组。

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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 *** **@gmailwrites： pereges <Br*****@gmailwrites:

< 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< www.cpax.uk> 电子邮件：-http：// www。 + rjh @ 谷歌用户：< www.cpax.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 <www.cpax.uk> Email: -www. +rjh@ Google users: <www.cpax.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 ***** @ gmailwrites： >pereges <Br*****@gmailwrites:

< 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 < www.ghoti/~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 <www.ghoti/~kst> Nokia "We must do something. This is something. Therefore, we must do this." -- Antony Jay and Jonathan Lynn, "Yes Minister"