C 对一个数组排序并将顺序复制到另一个数组
我有两个并排排列的数组,一个列出不同的队,另一个列出分数。我能按分数的降序排列。然后,可以使用此命令将相应的团队移动到领导委员会的正确位置吗?将两支得分为100分的球队(美国队和德国队)排在榜首C 对一个数组排序并将顺序复制到另一个数组,c,arrays,C,Arrays,我有两个并排排列的数组,一个列出不同的队,另一个列出分数。我能按分数的降序排列。然后,可以使用此命令将相应的团队移动到领导委员会的正确位置吗?将两支得分为100分的球队(美国队和德国队)排在榜首 #include <stdio.h> int main() { char teams[18][20]={"England","Ireland","Wales","Scotland","France","Italy","Germany","Uraguay","Belgium","USA"
#include <stdio.h>
int main()
{
char teams[18][20]={"England","Ireland","Wales","Scotland","France","Italy","Germany","Uraguay","Belgium","USA","Mexico","Australia","Belize","Denmark","Sweden","Japan","South Africa","Algeria"};
int points[18]={43,5,77,23,89,0,100,46,94,100,45,55,32,65,11,37,26,78};
int i;
int j;
int a;
for (i = 0; i < 18; ++i)
{
printf("%i ",i+1);
printf("%s",teams[i]);
printf("\t%d\n", points[i]);
}
printf("\n");
for (i = 0; i < 18; ++i)
{
for (j = i + 1; j < 18; ++j)
{
if (points[i] < points[j])
{
a = points[i];
points[i] = points[j];
points[j] = a;
}
}
}
for (i = 0; i < 18; ++i)
{
printf("%i ",i+1);
printf("%s",teams[i]);
printf("\t%d\n", points[i]);
}
return 0;
}
#包括
int main()
{
char团队[18][20]={“英格兰”、“爱尔兰”、“威尔士”、“苏格兰”、“法国”、“意大利”、“德国”、“乌拉瓜”、“比利时”、“美国”、“墨西哥”、“澳大利亚”、“伯利兹”、“丹麦”、“瑞典”、“日本”、“南非”、“阿尔及利亚”};
整数点[18]={43,5,77,23,89,0100,46,94100,45,55,32,65,11,37,26,78};
int i;
int j;
INTA;
对于(i=0;i<18;++i)
{
printf(“%i”,i+1);
printf(“%s”,团队[i]);
printf(“\t%d\n”,点[i]);
}
printf(“\n”);
对于(i=0;i<18;++i)
{
对于(j=i+1;j<18;++j)
{
如果(点[i]<点[j])
{
a=点[i];
点[i]=点[j];
点[j]=a;
}
}
}
对于(i=0;i<18;++i)
{
printf(“%i”,i+1);
printf(“%s”,团队[i]);
printf(“\t%d\n”,点[i]);
}
返回0;
}
如评论中所述,典型的解决方案是将数据建模为结构数组,而不是单独的数组。这是有意义的,因为数据是相互关联的
你会有类似于:
struct score {
const char *name;
int points;
} scores[] = {
{ "England", 43 },
{ "Ireland", 5 },
/* and so on */
};
然后,您可以使用
qsort()
(或者您自己的排序代码,如果您感兴趣的话)对整个结构实例进行排序,因为整个结构都在移动,所以所有数据都将保持在一起。如注释中所述,典型的解决方案是将数据建模为结构数组,而不是单独的数组。这是有意义的,因为数据是相互关联的
你会有类似于:
struct score {
const char *name;
int points;
} scores[] = {
{ "England", 43 },
{ "Ireland", 5 },
/* and so on */
};
然后,您可以使用
qsort()
(或者您自己的排序代码,如果您感兴趣的话)对整个结构实例进行排序,因为整个结构都在移动,所以所有的数据都将保持在一起。在排序时也会排列您的团队数组
a = points[i];
b = teams[i];
points[i] = points[j];
teams[i] = teams[j];
points[j] = a;
teams[j] = b;
在排序时,还要安排您的团队阵列
a = points[i];
b = teams[i];
points[i] = points[j];
teams[i] = teams[j];
points[j] = a;
teams[j] = b;
显而易见的方法(正如其他人所指出的)是将数组嵌入到结构中,但如果您被迫使用并行数组,您可以构建自己的函数并同时对两个数组进行排序:
#include <stdio.h>
static int comp(const void *a, const void *b)
{
return *(int *)a - *(int *)b;
}
static void swap(int v1[], char *v2[], int a, int b)
{
int temp1;
char *temp2;
temp1 = v1[a];
v1[a] = v1[b];
v1[b] = temp1;
temp2 = v2[a];
v2[a] = v2[b];
v2[b] = temp2;
}
static void sort(int v1[], char *v2[], int left, int right, int (*comp)(const void *, const void *))
{
int i, last;
if (left >= right) return;
swap(v1, v2, left, (left + right) / 2);
last = left;
for (i = left + 1; i <= right; i++) {
if (comp(&v1[i], &v1[left]) < 0)
swap(v1, v2, ++last, i);
}
swap(v1, v2, left, last);
sort(v1, v2, left, last - 1, comp);
sort(v1, v2, last + 1, right, comp);
}
int main(void)
{
char *teams[] = {"England","Ireland","Wales","Scotland","France","Italy","Germany","Uraguay","Belgium","USA","Mexico","Australia","Belize","Denmark","Sweden","Japan","South Africa","Algeria"};
int points[] = {43,5,77,23,89,0,100,46,94,100,45,55,32,65,11,37,26,78};
size_t i, n = sizeof(points) / sizeof(*points);
sort(points, teams, 0, n - 1, comp);
for (i = 0; i < n; i++) {
printf("%s->%d\n", teams[i], points[i]);
}
return 0;
}
显而易见的方法(正如其他人所指出的)是将数组嵌入到结构中,但如果您被迫使用并行数组,您可以构建自己的函数并同时对两个数组进行排序:
#include <stdio.h>
static int comp(const void *a, const void *b)
{
return *(int *)a - *(int *)b;
}
static void swap(int v1[], char *v2[], int a, int b)
{
int temp1;
char *temp2;
temp1 = v1[a];
v1[a] = v1[b];
v1[b] = temp1;
temp2 = v2[a];
v2[a] = v2[b];
v2[b] = temp2;
}
static void sort(int v1[], char *v2[], int left, int right, int (*comp)(const void *, const void *))
{
int i, last;
if (left >= right) return;
swap(v1, v2, left, (left + right) / 2);
last = left;
for (i = left + 1; i <= right; i++) {
if (comp(&v1[i], &v1[left]) < 0)
swap(v1, v2, ++last, i);
}
swap(v1, v2, left, last);
sort(v1, v2, left, last - 1, comp);
sort(v1, v2, last + 1, right, comp);
}
int main(void)
{
char *teams[] = {"England","Ireland","Wales","Scotland","France","Italy","Germany","Uraguay","Belgium","USA","Mexico","Australia","Belize","Denmark","Sweden","Japan","South Africa","Algeria"};
int points[] = {43,5,77,23,89,0,100,46,94,100,45,55,32,65,11,37,26,78};
size_t i, n = sizeof(points) / sizeof(*points);
sort(points, teams, 0, n - 1, comp);
for (i = 0; i < n; i++) {
printf("%s->%d\n", teams[i], points[i]);
}
return 0;
}
将团队和点放在结构数组中,而不是放在单独的数组中。将团队和点放在结构数组中,而不是放在单独的数组中。