用于排列的 C++ 递归算法 [英] C++ recursive algorithm for permutation

问题描述

permute() 函数陷入无限循环，我似乎找不到原因?我尝试通过删除递归调用来检查函数，它似乎工作正常.我也有基本情况，所以不知道问题出在哪里.

The permute() function runs into an infinite loop and I can't seem to find why? i tried checking the function by removing the recursive call and it seems to be working fine. I also have the base case, so don't know where is the problem.

#include <iostream>
#include <string>
#include <vector>

using namespace std;

string smallString(string s, int k){    // computes a string removing the character at index k 
    int i,j;
    string res;
    for(i=0,j=0;j<s.length();i++,j++){
        if(i==k){j++;}
        res.push_back(s[j]);
    }
return res;
}
void permute(string s1, string s2, size_t len){
    if(len==1)
        {cout<<"length is equal to 1"<<(s1+s2)<<'\n'; return;} //base case
    else{
        for(int i =0;i<len;i++){
            string temp= s2.substr(i,1);
            s1.append(temp);
            string fin = smallString(s2,i);
            //cout<<temp<<'\t'<<s1<<'\t'<<fin<<'\t'<<fin.length()<<'\n';
            permute(s1,fin,fin.length());
            s1.erase((s1.length()-1));
            //cout<<"printing s1 : "<<s1<<'\n';
        }
    }
}

int main(){
    string s2="abc";
    string s1="";
    permute(s1,s2,s2.length());
    return 0;
}

推荐答案

您的问题似乎出在smallString"函数上.在该函数中，OutOfRange 用于 s[j].我把打印像

Your problem seems to be in the "smallString" function. In that function, OutOfRange is used in s[j]. I put print like

for(i=0,j=0;j<s.length();i++,j++)
{
    if(i==k){j++;}
      cout<<"smallString:: s ::"<<s<<'\t'<<"k::"<<k<<'\n';
    res.push_back(s[j]); //Problem is here...
}

现在输出打印就像

smallString:: s ::abc k::0

smallString:: s ::abc k::0

smallString:: s ::abc k::0

smallString:: s ::abc k::0

smallString:: s ::bc k::0

smallString:: s ::bc k::0

smallString:: s ::bc k::1

smallString:: s ::bc k::1

smallString:: s ::bc k::1

smallString:: s ::bc k::1

smallString::s ::b k::0

smallString:: s ::b k::0

smallString:: s ::b k::1

smallString:: s ::b k::1

smallString:: s ::b k::1..

smallString:: s ::b k::1 . .

因此，在某个时间点它是s ::b k::1"，因此您要从字符串b"中选择位置 1 处的字符.

So, At one point of time it comes as "s ::b k::1" so you are selecting the character at the position 1 from the string "b".

String 基本上是一个从 0 到 (n-1) 的字符数组.对于字符串b"，只有第 0 个位置具有字符b".但我们正在访问一个不存在的职位.

String is basically an char array which start from 0 to (n-1). For string "b" only 0th position is having character 'b'. but we are accessing an non-existing position.

所以它抛出错误并从这里开始连续循环.

So it throws error and start looping continuously from here.

解决您的问题:

for(i=0,j=0;j<s.length();i++,j++)
{
    if(i==k)
    {
        j++;
    }
    else
    {
        cout<<"smallString:: s ::"<<s<<'\t'<<"k::"<<k<<'\n';
        res.push_back(s[j]);
    }
}

这篇关于用于排列的 C++ 递归算法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！