Question

如何从字符串中删除模式的所有实例？

string str = "red tuna, blue tuna, black tuna, one tuna";
string pattern = "tuna";

Answer 1

从字符串

中删除模式的所有实例

#include <string>
#include <iostream>

using namespace std;

void removeSubstrs(string& s, string& p) { 
  string::size_type n = p.length();
  for (string::size_type i = s.find(p);
      i != string::npos;
      i = s.find(p))
      s.erase(i, n);
}

int main() {

  string str = "red tuna, blue tuna, black tuna, one tuna";
  string pattern = "tuna";

  removeSubstrs(str, pattern);
  cout << str << endl;
}

Answer 2

这是一个基本问题，您最好查看标准库中的string功能。

经典解决方案

#include <iostream>
#include <string>

int main() { 
   std::string str = "red tuna, blue tuna, black tuna, one tuna";
   std::string pattern = "tuna";

   std::string::size_type i = str.find(pattern);
   while (i != std::string::npos) {
     str.erase(i, pattern.length());
     i = str.find(pattern, i);
   }

   std::cout << str;
}

Example

RegEx解决方案

从C ++ 11开始，你有另一种解决方案（感谢Joachim提醒我这一点）基于regular expressions

#include <iostream>
#include <string>
#include <regex>

int main() { 
   std::string str = "red tuna, blue tuna, black tuna, one tuna";
   std::regex pattern("tuna");

   std::cout << std::regex_replace(str, pattern, "");
}

Example

Answer 3

尝试类似：

void replaceAll(std::string& str, const std::string& from, const std::string& to) {
    if(from.empty())
        return;
    size_t start_pos = 0;
    while((start_pos = str.find(from, start_pos)) != std::string::npos) {
        str.replace(start_pos, from.length(), to);
        start_pos += to.length(); // In case 'to' contains 'from', like replacing 'x' with 'yx'
    }
}

来自Replace part of a string with another string

Answer 4

这是更好理解的基本逻辑 这是c ++中的代码

    string s,x;                //s is the string , x is the substring
    int a,l; 
    cin>>s>>x;
    l=x.length();
    while(true)
    {
        a=s.find(x);
        if(a==-1)
        {break;}                       // if substring is not found
        else
        {
        s.erase(a,l);                  //if substring is found 
        }
    }
    if(s.length()==0)                  // if string length becomes null printing 0 
        cout<<0<<"\n";      
    else 
        cout<<s<<endl;                 // else printing the string

示例：输入shahaha 输出shha

Answer 5

一种更快的算法，一次构建一个字符并检查结尾是否与子字符串匹配。以下在 o(substring * string) 中运行，而上述解决方案在 o(s^2/t) 中运行。

string S, T;
  cin >> S >> T;

  /* Build R, the result string, one character at a time. */
  string R;
  for (int i = 0; i < S.size(); i++) {
    R += S[i];

    /* If the end of R matches T then delete it. */
    if (R.size() >= T.size() && R.substr(R.size() - T.size()) == T) {
      R.resize(R.size() - T.size());
    }
  }

  cout << R << endl;

如何从字符串中删除所有子字符串

5 个答案:

经典解决方案

RegEx解决方案