如何在 C++ 中创建多个向量的组合而无需硬编码循环?

2021-12-16 00:00:00 递归 algorithm combinations vector c++

我有几个看起来像这样的数据:

I have several data that looks like this:

Vector1_elements = T,C,A
Vector2_elements = C,G,A
Vector3_elements = C,G,T
..... up to ...
VectorK_elements = ...

#Note also that the member of each vector is always 3.

我想要做的是在 Vector1 到 VectorK 中创建所有元素组合.因此最终我们希望得到这个输出(使用 Vector1,2,3):

What I want to do is to create all combination of elements in Vector1 through out VectorK. Hence in the end we hope to get this output (using Vector1,2,3):

TCC
TCG
TCT
TGC
TGG
TGT
TAC
TAG
TAT
CCC
CCG
CCT
CGC
CGG
CGT
CAC
CAG
CAT
ACC
ACG
ACT
AGC
AGG
AGT
AAC
AAG
AAT

我现在遇到的问题是我的以下代码通过对循环进行硬编码来实现.由于向量的数量可以变化，我们需要一种灵活的方法来获得相同的结果.有吗?

The problem I am having now is that the following code of mine does that by hardcoding the loops. Since number of Vectors can be varied, we need a flexible way to get the same result. Is there any?

我的这段代码最多只能处理 3 个向量(硬编码):

This code of mine can only handle up to 3 Vectors (hardcoded):

#include <iostream>
#include <vector>
#include <fstream>
#include <sstream>
using namespace std;


int main  ( int arg_count, char *arg_vec[] ) {

    vector <string> Vec1;
          Vec1.push_back("T");
          Vec1.push_back("C");
          Vec1.push_back("A");

    vector <string> Vec2;
          Vec2.push_back("C");
          Vec2.push_back("G");
          Vec2.push_back("A");

    vector <string> Vec3;
          Vec3.push_back("C");
          Vec3.push_back("G");
          Vec3.push_back("T");



     for (int i=0; i<Vec1.size(); i++) {
        for (int j=0; j<Vec2.size(); j++) {
            for (int k=0; k<Vec1.size(); k++) {
                cout << Vec1[i] << Vec2[i] << Vec3[k] << endl;
            }
        }
     }



    return 0;
}

推荐答案

这样做可以解决问题:

void printAll(const vector<vector<string> > &allVecs, size_t vecIndex, string strSoFar)
{
    if (vecIndex >= allVecs.size())
    {
        cout << strSoFar << endl;
        return;
    }
    for (size_t i=0; i<allVecs[vecIndex].size(); i++)
        printAll(allVecs, vecIndex+1, strSoFar+allVecs[vecIndex][i]);
}

致电:

printAll(allVecs, 0, "");

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