字符串操作

字符串处理

1. 字符串去分割符

   void RemoveDelimiter(string& str)
   {
       array<char, 2> delimiterArray{',', ' '};
       for (int i = 0; i < str.size(); ++i){
           if (auto it = find(delimiterArray.begin(), delimiterArray.end(), str[i]); it != delimiterArray.end()){
           	str.erase(i, 1);
             	i--;
       	}
       }
   }
   

2. 字符串分割

   // str = "123+456=689"
   vector<string> SplitString(string& str)
   {
       array<char, 4> delimiterNums{'+', '-', '*', '/'};
       vector<string> splitedString;
       for (int i = 0; i < str.size(); ++i){
           if (auto it = find(delimiterNums.begin(), delimiterNums.end(), s[i]); it != str.end()){
               char delimiter = s[i];
               splitedString.push_back(str.substr(0, i));
               splitedString.push_back(string{delimiter});
               str.erase(0, i+1);
               i = -1;
           }
       }
       splitedString.push_back(str);
   }
   

3. 字符串去两边空格

   void Stripe(string& s)
   {
       for (int i = 0; i < s.size(); ++i){
           if (s[i] != ' '){
               s = s.substr(i);
               break;
           }
       }
       for (int i = s.size(); i >= 0; --i){
           if (s[i] != ' '){
               s = s.substr(0，i+1);
               break;
           }
       }
   }
   

4. 字符串加法

   字符串加法较为简单，直接将每一位进行相加然后即可，具体可分为四步。为了方便计算始终保证a字符串的长度更长。

   1. 判断长度，交换并反向
   2. 遍历取一位进行计算
   3. 处理剩余字符串，注意前面计算可能进位不为0
   4. 反向结果字符串

   // 大数加法 cyclomatic complexity : 5
   std::string StringAdd(std::string& a, std::string& b)
   {
       std::string res;
       // 交换并反向，然后取每一位进行加和
       if (a.size() < b.size()){
           swap(a, b);
       }
       reverse(a.begin(), a.end());
       reverse(b.begin(), b.end());
   
       // 取每一位进行加和
       int carry = 0;
       for (int i = 0; i < b.size(); ++i){
           int sum = (a[i] - '0') + (b[i] - '0') + carry;
           res.push_back(sum%10 + '0');
           carry = sum/10;
       }
   
       // 处理更长的剩余字符串
       int i = b.size();
       while (i != a.size()){
           if (carry > 0){
               int sum = a[i] + carry;
               res.push_back(sum%10);
               carry = sum/10;
           }else{
               res.push_back(a[i]);
           }
           i++;
       }
       reverse(res.begin(), res.end());
       return res;
   }