创建一组字符子集的最佳解决方案是什么?

你想要 power set . 这是可以计算出来的 recursively and inductively . ;-)

或者,长度为n的字符串有2^n个子集。所以写两个嵌套循环:i从0计数到2^n-1(对于子集),j从0计数到n-1(对于第i个子集中的字符)。当且仅当i的第j位为1时,输出字符串的第j个字符。

(嗯,你确实说过“最好”是主观的…)

    char str [] = "abc";
    int n = strlen(str); // n is number of elements in your set

    for(int i=0; i< (1 << n); i++) { // (1 << n) is equal to 2^n
        for(int j=0; j<n; j++) { // For each element in the set
            if((i & (1 << j)) > 0) { // Check if it's included in this subset. (1 << j) sets the jth bit
                cout << str[j];
            }
        }
        cout << endl;
    }

def subsets(s):
    r = []
    a = [False] * len(s)
    while True:
        r.append("".join([s[i] for i in range(len(s)) if a[i]]))
        j = 0
        while a[j]:
            a[j] = False
            j += 1
            if j >= len(s):
                return r
        a[j] = True

print subsets("abc")

请原谅伪代码。。。

int i = 0;
Results.push({});

While(i > Inset.Length) {
   Foreach(Set s in Results) {
    If(s.Length == i) {
       Foreach(character c in inSet)
          Results.push(s+c);
    }
    i++;
}

//recursive solution in C++
set<string> power_set_recursive(string input_str)
{
    set<string> res;
    if(input_str.size()==0) {
        res.insert("");
    } else if(input_str.size()==1) {
        res.insert(input_str.substr(0,1));
    } else {
        for(int i=0;i<input_str.size();i++) {
            set<string> left_set=power_set_iterative(input_str.substr(0,i));
            set<string> right_set=power_set_iterative(input_str.substr(i,input_str.size()-i));
            for(set<string>::iterator it1=left_set.begin();it1!=left_set.end();it1++) {
                for(set<string>::iterator it2=right_set.begin();it2!=right_set.end();it2++) {
                    string tmp=(*it1)+(*it2);
                    sort(tmp.begin(),tmp.end());
                    res.insert(tmp);
                }
            }
        }
    }
    return res;
}


//iterative solution in C++
set<string> power_set_iterative(string input_str)
{
    set<string> res;
    set<string> out_res;
    res.insert("");
    set<string>::iterator res_it;
    for(int i=0;i<input_str.size();i++){
        for(res_it=res.begin();res_it!=res.end();res_it++){
                string tmp=*res_it+input_str.substr(i,1);
                sort(tmp.begin(),tmp.end());
                out_res.insert(tmp);
        }
        res.insert(input_str.substr(i,1));
        for(set<string>::iterator res_it2=out_res.begin();res_it2!=out_res.end();res_it2++){
            res.insert(*res_it2);
    }
    out_res.clear();
    }
    return res;
}