Question

我将如何解决这个问题？我想在字符串的所有位置搜索一个大字符串。

Answer 1

使用std::string::find。你可以这样做：

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，Nawaz！更好？

Answer 2

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算法可以找到O（N）复杂度的所有子串。

修改

还有另一个算法，所谓的“Z函数”具有O（N）复杂度，但我找不到这个算法的英文源（也许是因为还有另一个更有名的东西同名 - Z- Riman的功能），所以只需将其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 3

我将添加完整性，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);
}

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

Answer 4

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

Here是文档。

以下是本文档中的示例：

// 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;
}

如何在字符串中找到子字符串的所有位置？

4 个答案: