Create a full array of 1~n

Enter a positive integer n, which is the full rank of the output n.

Sample Input 1:

3

Sample Output 1:

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1

Analysis:

Output all permutations in sequence from small to large in dictionary order.

(Dictionary order: the dictionary order size relationship of two sequences is equivalent to the size relationship at the first different position from the beginning of a sequence)

Using array A to save the number in the arrangement, the set S represents the remaining number.

Then there are

Method 1:

1. Pseudo-code:

int Dfs (A,S,CUR) {//finds the CUR number in the arrangement.

If (s== null) outputs this arrangement;

Else

Take the element I in the set S;

a[cur]=i;//save in a array

DFS (A,S,CUR+1);

}

s set can be omitted, directly denoted by a, if I exist in a[1]~a[cur-1], then I have been used, otherwise the description I in the set S.

Because you want to start with the first number, you first call DFS (1)//cur=1

Source:

#include <cstring>

using namespace Std;

int a[100],s, n;;

void dfs (int cur) {///function has no return value, so the type is void

if (cur==n+1) {

for (int i=1;i<n;i++) cout<<a[i]<< "";

cout<<a[n]<<endl;

s++;

}

else for (int i=1;i<=n;i++) {

int ok=1;

for (int j=1;j<cur; j + +)//Because we are currently looking for the cur number, I compare it with the cur-1 number I have found earlier

if (a[j]==i) ok=0;

if (OK) {a[cur]=i;d fs (cur+1);}

}

}

int main () {

memset (A,0,sizeof (a));

cin>>n;

DFS (1);

cout<<s<<endl;

return 0;

}

Thinking:

(1) Global variables and local variables;

For example a array and variable n,s are used in both main and DFS, so defined as a global variable, cur is only used in the currently called function, so it is defined as a local variable.

(2) How the function does not return a value when it is processed.

function has no return value, the function type is void, can have no return statement, or can have, but there is no expression after

return;

2.

Pseudo code:

int DFS (cur) {//has found cur numbers and is ready to find the next one.

If (s== null) outputs this arrangement;

Else

Take the element I in the set S;

a[cur+1]=i;//save in a array

DFS (cur+1);

}

0 numbers were found initially, so Dfs is called first (0)

1#include <iostream>2#include <cstring>3 using namespacestd;4 inta[ -],s=0, N;5 voidDfsintcur) {//function has no return value when type is void6     if(cur==N) {7          for(intI=1; i<n;i++) cout<<a[i]<<" ";8cout<<a[n]<<Endl;9s++;Ten     } One     Else A           for(intI=1; i<=n;i++){ -               intok=1; -                for(intj=1; j<=cur;j++) the                     if(a[j]==i) ok=0; -               if(OK) { -a[cur+1]=i;  -DFS (cur+1);  +             } -           }       +}

View Code

1#include <iostream>2#include <cstring>3 using namespacestd;4 inta[ -],s, n;;5 voidDfsintcur) {6     if(cur==N) {7          for(intI=1; i<n;i++) cout<<a[i]<<" ";8cout<<a[n]<<Endl;9s++;Ten     return; One     } A       for(intI=1; i<=n;i++){ -           intok=1; -            for(intj=1; j<=cur;j++) the                 if(a[j]==i) ok=0; -           if(OK) { -a[cur+1]=i;  -DFS (cur+1);  +         } -      }       +}

View Code

Create a full array of 1~n