如何找到字符串中子字符串的所有位置？

Question

我想在一个大字符串中搜索一个字符串的所有位置。

Answer 1

另外两个答案是正确的，但它们非常慢并且具有 O(N^2) 复杂度。但是有 Knuth-Morris-Pratt 算法，它可以找到 O(N) 复杂度的所有子字符串。

编辑：

此外，还有另一种算法：所谓的“Z 函数”，复杂度为 O(N)，但我找不到该算法的英文来源（可能是因为还有另一个更著名的算法） name - Rieman 的 Z 函数），所以我将把它的代码放在这里并解释它的作用。

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
}

Answer 2

使用std::string::find。您可以执行以下操作：

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;
}

---

编辑：Doh！谢谢你的评论，纳瓦兹！更好？

Answer 3

为了完整起见，我补充一下，std::search 可以使用另一种方法，类似于std::string::find，不同之处在于您使用迭代器，例如：

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);
}

我发现这个有时优于std::string::find，尤其是。如果您将字符串表示为vector<char>。

Answer 4

只需use std::string::find()，它返回找到子字符串的位置，如果没有找到，则返回std::string::npos。

Here 是文档。

这里是取自本文档的示例：

// 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;
}