c - 查找数组中n个数字的所有可能排列

我有一个数组，它包含如下内容[25,15,8,20]
我想找到所有可能的数字安排。
预期产量：

25 15 8 20
25 15 20 8
25 20 15 8
25 20 8 15
25 8 20 15
25 8 15 20
15 25 8 20
15 25 20 8
15 20 25 8
15 20 8 25
15 8 20 25
15 8 25 20
20 25 15 8
20 25 8 15
20 8 25 15
20 8 15 25
20 15 25 8
20 15 8 25
8 15 20 25
8 15 25 20
8 25 15 20
8 25 20 15
8 20 15 25
8 20 25 15


void print(int *num, int n)
{
 int i;
 for ( i = 0 ; i < n ; i++)
    printf("%d ", num[i]);
   printf("\n");
}
int main()
{
 int num[N];
 int *ptr;
 int temp;
 int i, n, j;
 printf("\nHow many number you want to enter: ");
    scanf("%d", &n);
 printf("\nEnter a list of numbers to see all combinations:\n");
 for (i = 0 ; i < n; i++)
    scanf("%d", &num[i]);
 for (j = 1; j <= n; j++) {
    for (i = 0; i < n-1; i++) {
        temp = num[i];
        num[i] = num[i+1];
        num[i+1] = temp;
        print(num, n);
 }
}
 return 0;
}

上面的程序没有给出所有可能的输出如何获得内部交换和组合

最佳答案

诀窍是，不能只排列相邻的值，当3固定在index=1时，4永远不会放在一边，所以必须逐步将排列扩展到更多的值。

#include <stdio.h>
#include <string.h>
#include <malloc.h>

void print(int *num, int n)
{
 int i;
 for ( i = 0 ; i < n ; i++)
    printf("%d ", num[i]);
   printf("\n");
}

int*permute(int*i,int h)
{
     int temp = *i;
    *i = *(i+h);
    *(i+h) = temp;
return i+1;

}
void recursive_permute(int*i,int *j,int n)
{
    if((j-i)==n-1) {print(i,n);return;};

    int *tmparray=(int*)malloc(n*sizeof(int));
    memcpy(tmparray,i,n*sizeof(int));
    recursive_permute(tmparray,tmparray+(j-i+1),n);

    for (int h=1;h<n-(j-i);h++) recursive_permute(tmparray,permute(tmparray+(j-i),h),n);
}

int main()
{
 int num[100];
 int *ptr;
 int temp;
 int i, n, j;
 printf("\nHow many number you want to enter: ");
    scanf("%d", &n);
 printf("\nEnter a list of numbers to see all combinations:\n");
 for (i = 0 ; i < n; i++)
    scanf("%d", &num[i]);
 printf("my recursive method ---------------------------\n");
 recursive_permute(num,num,n);

 printf("your method -----------------------------------\n");

 for (j = 1; j <= n; j++) {
    for (i = 0; i < n-1; i++) {
        temp = num[i];
        num[i] = num[i+1];
        num[i+1] = temp;
        print(num, n);
 }
}
 return 0;
}

See it running here