C + + output full permutation recursive algorithm detailed explanation

Central:
Set R={R1,R2,..., rn} is the n element to be arranged, Ri=r-{ri}.
Perm (x) represents the arrangement of the prefix ri, preceded by each permutation of the fully arranged Perm (x).
(1) When N=1, Perm (r) = (R), where r is the only element in set R;
(2) When n>1, Perm (R) can be composed of (R1) +perm (R1), (R2) +perm (R2),..., (RN) +perm (RN).

So how do we do it? Let's take a practical example, assuming there's a sequence of 1,2,3,4
So the 1,2,3,4 of the whole arrangement
Perm ({1,2,3,4}) =1perm ({2,3,4}) +2perm ({1,3,4}) +3perm ({1,2,4}) +4perm (+/-)
So we just have to exchange each number, and the first number is not to get the next sequence?
For example {1,2,3,4} The first and second number exchange, then do not get 2 {1,3,4}, then we use a practical example of how the program is running

Specific algorithm flow:

Sequence: {x-man} first swap with first
You can get 1 {2,3} to put the sequence {2,3} into the perm function recursively, and then

--Recursive {2,3}
Sequence {2,3} First interchange with first
Get 2{3}, Output Low=high (at this point, because the sequence {3} has only one digit, so the output list)
The sequence {2,3} is first swapped back with the first, and the result is still {2,3}
Sequence {2,3} First and second swap
Get 3{2}, Output 1,3,2
{3,2} again first swap back with second, change back to {2,3}
-–{2,3} recursive complete sequence restitution

Sequence: {x-man} First and second swap
Can get 2,{1,3}
--Recursive {1,3}
Sequence {1,3} First interchange with first
Get 1{3}, Output 2,1,3
The sequence {1,3} is first swapped back with the first, and the result is still {1,3}
Sequence {1,3} First and second swap
Get 3{1}, Output 2,3,1
{3,1} again first swap back with second, change back to {1,3}
-–{1,3} recursion complete
Sequence {2,1,3} First and second interchange
Sequence restitution {.

Series: {X-man} First and third Exchange
Can get 3,{1,2}
--Recursive {+}
Sequence {x} First and first swap
Get 1{2}, Output 3,1,2
The first exchange returned with the first one, and the result is still {
Sequence {x} First and second swap
Get 2{1}, Output 3,2,1
{2,1} again first with the second exchange back, change back to {
-–{1,2} recursion complete
Sequence {3,1,2} First and second interchange
Sequence restitution {.

The algorithm can be written in simple
Perm ({=1perm}) ({2,3}) +2perm ({1,3}) +3perm ({.})
Perm ({2,3}) =2perm ({3}) +3perm ({2})
Perm ({1,3}) =1perm ({3}) +3perm ({1})
Perm ({)}) =1perm ({2}) +2perm ({1})

C + + code:

#include <iostream>using namespace STD;voidSwapint&a,int&AMP;B) {intTemp=a;    A=b; B=temp;}voidPermint List[],intLowintHigh) {if(Low==high) {//When Low==high, the list is one of the permutations, the output list         for(intI=0; i<=low;i++)cout<<List[i];cout<<endl; }Else{ for(inti=low;i<=high;i++) {//Each element is exchanged with the first elementSwapList[I],List[Low]); PermList, low+1, high);//Exchange, get sub-sequence, using function perm to get the whole sequence of sub-sequencesSwapList[I],List[Low]);///Finally, swap elements back, restore, and then swap another element}    }}intMain () {int List[]={1,2,3};p ERM (List,0,2);return 0;}

Program results:

123132213231321312

C + + output full permutation recursive algorithm detailed explanation