Question

任何人都可以完成以下程序吗？

感谢。

void RemoveDuplicates (string& input)

{    
    string nonRepeatedChars (input);

    sort(nonRepeatedChars.begin(), nonRepeatedChars.end());
    //cout << nonRepeatedChars <<endl;
    string::iterator it = unique(nonRepeatedChars.begin(), nonRepeatedChars.end());
    //cout << nonRepeatedChars <<endl;
    nonRepeatedChars.erase(it, nonRepeatedChars.end());
    cout << "nonRepeatedChars = "<< nonRepeatedChars <<endl;

    for(string::iterator i = input.begin(); i != input.end(); i++)
    {
        cout << "*i = " << *i <<endl;
        size_t found = nonRepeatedChars.find(*i);
        cout << "found = "<< found <<endl;
        if (found != string::npos)
        {
             input.erase(i);
             cout << "input = " << input <<endl;
        }
        else
        {
            nonRepeatedChars.erase(found, 1);
            cout << "nonRepeatedChars = "<< nonRepeatedChars <<endl;
        }
    }

    cout << "Final Input = " << input <<endl;
}

Answer 1

解决方案一如既往：

std::string RemoveDuplicates (const std::string& input) {
  std::string newT(input);
  std::string::iterator it = std::unique(newT.begin(), newT.end());
  newT.erase(it, newT.end());
  return newT;
}

另一种返回新字符串的解决方案：

std::string RemoveDuplicates (const std::string& input) {
  std::string newInput;
  const char * prev = nullptr;
  for (const auto & ch : input) {
    if (!prev || (*prev != ch)) {
      newInput.push_back(ch);
    }
    prev = &ch;
  }
  return newInput;
}

如果需要的结果是你好 - ＆gt; helo 然后解决方案是：

std::string RemoveDuplicates (const std::string& input) {
  std::string newInput;
  std::set<char> addedChars;
  for (const auto & ch : input) {
    if (addedChars.end() == addedChars.find(ch)) {
      newInput.push_back(ch);
      addedChars.insert(ch);
    }
  }
  return newInput;
}

如果您需要保存字符顺序并删除重复项：

Theme.Material

从字符串中删除所有重复的字符（STL）

1 个答案: