开发者

How do I find all the positions of a substring in a string?

开发者 https://www.devze.com 2023-03-01 01:31 出处:网络
I want to search a large string for all the locations of a str开发者_如何转开发ing.The two other answers are correct but they are very slow and have O(N^2) complexity. But there is the Knuth-Morris-Pr

I want to search a large string for all the locations of a str开发者_如何转开发ing.


The two other answers are correct but they are very slow and have O(N^2) complexity. But there is the Knuth-Morris-Pratt algorithm, which finds all substrings in O(N) complexity.

Edit:

Also, there is another algorithm: the so-called "Z-function" with O(N) complexity, but I couldn't find an English source for this algorithm (maybe because there is also another more famous one with same name - the Z-function of Rieman), so I will just put its code here and explain what it does.

void calc_z (string &s, vector<int> & z)
{
    int len = s.size();
    z.resize (len);

    int l = 0, r = 0;
    for (int i=1; i<len; ++i)
        if (z[i-l]+i <= r)
            z[i] = z[i-l];
        else
        {
            l = i;
            if (i > r) r = i;
            for (z[i] = r-i; r<len; ++r, ++z[i])
                if (s[r] != s[z[i]])
                    break;
            --r;
        }
}

int main()
{
    string main_string = "some string where we want to find substring or sub of string or just sub";
    string substring = "sub";
    string working_string = substring + main_string;
    vector<int> z;
    calc_z(working_string, z);

    //after this z[i] is maximal length of prefix of working_string
    //which is equal to string which starting from i-th position of
    //working_string. So the positions where z[i] >= substring.size()
    //are positions of substrings.

    for(int i = substring.size(); i < working_string.size(); ++i)
        if(z[i] >=substring.size())
            cout << i - substring.size() << endl; //to get position in main_string
}


Using std::string::find. You can do something like:

std::string::size_type start_pos = 0;
while( std::string::npos != 
          ( start_pos = mystring.find( my_sub_string, start_pos ) ) )
{
    // do something with start_pos or store it in a container
    ++start_pos;
}

EDIT: Doh! Thanks for the remark, Nawaz! Better?


I'll add for completeness, there is another approach that is possible with std::search, works like std::string::find, difference is that you work with iterators, something like:

std::string::iterator it(str.begin()), end(str.end());
std::string::iterator s_it(search_str.begin()), s_end(search_str.end());

it = std::search(it, end, s_it, s_end);

while(it != end)
{
  // do something with this position..

  // a tiny optimisation could be to buffer the result of the std::distance - heyho..
  it = std::search(std::advance(it, std::distance(s_it, s_end)), end, s_it, s_end);
}

I find that this sometimes outperforms std::string::find, esp. if you represent your string as a vector<char>.


Simply use std::string::find() which returns the position at which the substring was found, or std::string::npos if none was found.

Here is the documentation.

An here is the example taken from this documentation:

// string::find
#include <iostream>
#include <string>
using namespace std;

int main ()
{
  string str ("There are two needles in this haystack with needles.");
  string str2 ("needle");
  size_t found;

  // different member versions of find in the same order as above:
  found=str.find(str2);
  if (found!=string::npos)
    cout << "first 'needle' found at: " << int(found) << endl;

  found=str.find("needles are small",found+1,6);
  if (found!=string::npos)
    cout << "second 'needle' found at: " << int(found) << endl;

  found=str.find("haystack");
  if (found!=string::npos)
    cout << "'haystack' also found at: " << int(found) << endl;

  found=str.find('.');
  if (found!=string::npos)
    cout << "Period found at: " << int(found) << endl;

  // let's replace the first needle:
  str.replace(str.find(str2),str2.length(),"preposition");
  cout << str << endl;

  return 0;
}
0

精彩评论

暂无评论...
验证码 换一张
取 消