就是第一个数分别以后面的数进行交换
假设输入E = （a , b , c），
则 perm（E）= a.perm（b,c）+ b.perm（a,c）+ c.perm（a,b）
然后a.perm（b,c）= ab.perm（c）+ ac.perm（b）= abc + acb.依次递归进行
#include <stdio.h>
#define SWAP(x, y, t) ((t) = (x), (x) = (y), (y) = (t))
void perm(char *list, int i, int n);
int main(int argc, const char * argv[]) {
// insert code here...
char list[] = "abcd";
perm(list, 0, 3);
return 0;
}
void perm(char *list, int i, int n){
int j;
if(i == n){
for(j = 0; j <= n; j++){
printf("%c", list[j]);
}
printf(" ");
}
else{
for(j = i; j <= n; j++){
swap(list[i], list[j]);
perm(list, i + 1, n);
swap(list[i], list[j]);
}
}
}
具体递归流程如下：
***perm(abcd), 0, 3***
***perm(abcd),1, 3***
***perm(abcd),2, 3***
***perm(abcd),3, 3***
abcd
***perm(abdc),3, 3***
abdc
***perm(acbd),2, 3***
***perm(acbd),3, 3***
acbd
***perm(acdb),3, 3***
acdb
***perm(adcb),2, 3***
***perm(adcb),3, 3***
adcb
***perm(adbc),3, 3***
adbc
***perm(bacd),1, 3***
***perm(bacd),2, 3***
***perm(bacd),3, 3***
bacd
***perm(badc),3, 3***
badc
***perm(bcad),2, 3***
***perm(bcad),3, 3***
bcad
***perm(bcda),3, 3***
bcda
***perm(bdca),2, 3***
***perm(bdca),3, 3***
bdca
***perm(bdac),3, 3***
bdac
***perm(cbad),1, 3***
***perm(cbad),2, 3***
***perm(cbad),3, 3***
cbad
***perm(cbda),3, 3***
cbda
***perm(cabd),2, 3***
***perm(cabd),3, 3***
cabd
***perm(cadb),3, 3***
cadb
***perm(cdab),2, 3***
***perm(cdab),3, 3***
cdab
***perm(cdba),3, 3***
cdba
***perm(dbca),1, 3***
***perm(dbca),2, 3***
***perm(dbca),3, 3***
dbca
***perm(dbac),3, 3***
dbac
***perm(dcba),2, 3***
***perm(dcba),3, 3***
dcba
***perm(dcab),3, 3***
dcab
***perm(dacb),2, 3***
***perm(dacb),3, 3***
dacb
***perm(dabc),3, 3***
dabc
Program ended with exit code: 0
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