Solving the combination sequence

Today, we used the algorithm for solving the composite sequence and implemented it using C ++ and OCaml.

Problem description: There are arrays (linked lists, etc.) a, length len, and n elements are extracted from them to give all the removal sequences.

 

This is the problem of arrangement and combination. There are a total of C (len, n) sequences, which were once done by High School. The implementation process is actually relatively simple, as follows:

If the current position is s, a [s] can be obtained first, and n-1 elements are extracted from the remaining elements to form a sequence;

Or n elements are extracted from the remaining elements as a sequence instead of a [s ].

Iterate this process

C ++ code

#include<iostream>#include<vector>using namespace std;vector<vector<int> > pick(int l[],int s,int e,int n){vector<vector<int> > v;if(e-s+1>=n&&n>0){vector<vector<int> > res1 = pick(l,s+1,e,n-1);int size=res1.size();if(size>0){for(int i=0;i<size;i++){res1[i].push_back(l[s]);}}else{vector<int> tmp;tmp.push_back(l[s]);res1.push_back(tmp);}vector<vector<int> > res2 = pick(l,s+1,e,n);size=res2.size();for(int i=0;i<size;i++){res1.push_back(res2[i]);}v= res1;}return v;}int main(){int coe[]={1,2,3,4,5};vector<vector<int> > res = pick(coe,0,4,3);int size=res.size();for (int i=0;i<size;i++){vector<int> v=res[i];int len=v.size();for (int j=0;j<len;j++){cout<<v[j]<<",";}cout<<endl;}}

Running result

OCaml code

 1 let rec pick a s e len =
 2     let v = ref [] in
 3     if e-s+1>=len&&len>0 then
 4     begin
 5         let len0 = len in
 6         let res1 = pick a (s+1) e (len-1) in
 7         let len = List.length !res1 in
 8         let ele = List.nth a s in
 9         if len>0 then
10         begin
11             List.iter(fun r->
12                 r := ele::!r;
13             )!res1;
14         end else
15         begin
16             res1 := [ref [ele]];
17         end;
18         
19         let res2 = pick a (s+1) e len0 in
20         List.iter(fun r->
21             res1 := r::!res1;
22         )!res2;
23             
24         v := !res1;
25     end;
26         
27     v
28 in
29     
30 let a = [1;2;3;4;5] in
31 let res = pick a 0 4 3 in
32 
33 List.iter(fun r->
34     List.iter(fun r1->
35         Printf.printf "%d," r1;
36     )!r;
37     Printf.printf "\n";
38 )!res;

Running result