Question

我有一个包含让我们说[25,15,8,20]的数组我想找到所有可能的数字排列。

预期产出：

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;
}

上述程序未提供所有可能的输出。如何获得内部交换并获得组合

Answer 1

找到所有排列和排列数有几个方面。通过以下评论显示，要计算k元素总数中n个尺寸组的排列数，您会发现n上的阶乘除以{{1}的阶乘数可以组成k元素的大小组。对于查找四元素数组中所有4个元素的所有排列的情况，存在n个可能的排列。

然后递归地找到可用的排列。如果您有任何问题，请查看以下内容并告诉我：

使用/输出

#include <stdio.h> #include <stdlib.h> void swap (int *x, int *y); unsigned long long nfact (size_t n); unsigned long long pnk (size_t n, size_t k); void permute (int *a, size_t i, size_t n); void prnarray (int *a, size_t sz); int main (void) { int array[] = { 25, 15, 8, 20 }; size_t sz = sizeof array/sizeof *array; /* calculate the number of permutations */ unsigned long long p = pnk (sz , sz); printf ("\n total permutations : %llu\n\n", p); /* permute the array of numbers */ printf (" permutations:\n\n"); permute (array, 0, sz); putchar ('\n'); return 0; } /* Function to swap values at two pointers */ void swap (int *x, int *y) { int temp; temp = *x; *x = *y; *y = temp; } /* calculate n factorial */ unsigned long long nfact (size_t n) { if (n <= 0) return 1; unsigned long long s = n; while (--n) s *= n; return s; } /* calculate possible permutations */ unsigned long long pnk (size_t n, size_t k) { size_t d = (k < n ) ? n - k : 1; return nfact (n) / nfact (d); } /* permute integer array for elements 'i' through 'n' */ void permute (int *a, size_t i, size_t n) { size_t j; if (i == n) prnarray (a, n); else for (j = i; j < n; j++) { swap ((a+i), (a+j)); permute (a, i+1, n); swap ((a+i), (a+j)); // backtrack } } void prnarray (int *a, size_t sz) { size_t i; for (i = 0; i < sz; i++) printf (" %2d", a[i]); putchar ('\n'); }

注意：对于字符串，这种方法可以正常工作，但不会导致对可能的排列进行词汇排序。

Answer 2

诀窍是，你不能只置换相邻的值，看看当3被固定在index = 1时，4永远不会放弃它，所以你必须逐渐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;  
}

找到数组中n个数字的所有可能排列

2 个答案:

See it running here