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
}
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…