字符串函数优化？

Question

我是C++新手，我刚刚写了一个函数来告诉我字符串中的某些字符是否重复：

bool repeats(string s)
{
    int len = s.size(), c = 0;

    for(int i = 0; i < len; i++){
        for(int k = 0; k < len; k++){
            if(i != k && s[i] == s[k]){
                c++;
            }    
        }
    }
    return c;
}

...but我情不自禁地认为，对于它应该做的事情来说，它有点拥挤。有没有办法用更少的代码行来编写这样的函数？

Answer 1

有没有办法用更少的代码行来编写这样的函数？

使用std，您可能会这样做：

bool repeats(const std::string& s)
{
    return std::/*unordered_*/set<char>{s.begin(), s.end()}.size() != s.size();
}

Answer 2

#include <algorithm>

bool repeats(std::string s){
    for (auto c : s){
         if(std::count(s.begin(), s.end(), c) - 1)
             return true;
    }
    return false;
}

Answer 3

假设您不是在寻找重复子字符串：

#include <iostream>
#include <string>
#include <set>

std::set<char> ignore_characters{ ' ', '\n' };

bool has_repeated_characters(const std::string& input)
{
    // std::set<char> is a collection of unique characters
    std::set<char> seen_characters{};

    // loop over all characters in the input string
    for (const auto& c : input)
    {
        // skip characters to ignore, like spaces
        if (ignore_characters.find(c) == ignore_characters.end())
        {
            // check if the set contains the character, in C++20 : seen_characters.contains(c)
            // and maybe you need to do something with "std::tolower()" here too
            if (seen_characters.find(c) != seen_characters.end())
            {
                return true;
            }

            // add the character to the set, we've now seen it
            seen_characters.insert(c);
        }
    }

    return false;
}

void show_has_repeated_characters(const std::string& input)
{
    std::cout << "'" << input << "' ";

    if (has_repeated_characters(input))
    {
        std::cout << "has repeated characters\n";
    }
    else
    {
        std::cout << "doesn't have repeated characters\n";
    }
}

int main()
{
    show_has_repeated_characters("Hello world");
    show_has_repeated_characters("The fast boy"); 
    return 0;
}