I need to calculate permutations iteratively. The method signature looks like:
int[][] permute(int n)
For n = 3 for example, the r
The algorithm for stepping from one permutation to the next is very similar to elementary school addition - when an overflow occurs, "carry the one".
Here's an implementation I wrote in C:
#include
//Convenience macro. Its function should be obvious.
#define swap(a,b) do { \
typeof(a) __tmp = (a); \
(a) = (b); \
(b) = __tmp; \
} while(0)
void perm_start(unsigned int n[], unsigned int count) {
unsigned int i;
for (i=0; i0 && n[i-1] >= n[i]; i--);
tail = i;
if (tail > 0) {
/* Find the last item from the tail set greater than
the last item from the head set, and swap them.
Example: 0 3* (5 4* 2 1)
Becomes: 0 4* (5 3* 2 1) */
for (j=count-1; j>tail && n[j] <= n[tail-1]; j--);
swap(n[tail-1], n[j]);
}
/* Reverse the tail set's order */
for (i=tail, j=count-1; i