Difference between revisions of "Longest palindromic substring/lps.cpp"
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− | // Implementation of Jeuring's algorithm (see http://johanjeuring.blogspot.com/2007/08/finding-palindromes.html) | + | // Implementation of Jeuring's algorithm (see http://johanjeuring.blogspot.com/2007/08/finding-palindromes.html ) |
// Thanks to Frederick Akalin (see http://www.akalin.cx/2007/11/28/finding-the-longest-palindromic-substring-in-linear-time/ ) | // Thanks to Frederick Akalin (see http://www.akalin.cx/2007/11/28/finding-the-longest-palindromic-substring-in-linear-time/ ) | ||
// Brian Bi (bbi5291), 2010-11-14 | // Brian Bi (bbi5291), 2010-11-14 |
Revision as of 20:30, 14 November 2010
// Implementation of Jeuring's algorithm (see http://johanjeuring.blogspot.com/2007/08/finding-palindromes.html ) // Thanks to Frederick Akalin (see http://www.akalin.cx/2007/11/28/finding-the-longest-palindromic-substring-in-linear-time/ ) // Brian Bi (bbi5291), 2010-11-14 #include <iostream> #include <vector> #include <algorithm> using namespace std; template <class RAI1,class RAI2> void fastLongestPalindromes(RAI1 seq,RAI1 seqEnd,RAI2 out) { int seqLen=seqEnd-seq; int i=0,j,d,s,e,lLen,k=0; int palLen=0; while (i<seqLen) { if (i>palLen && seq[i-palLen-1]==seq[i]) { palLen+=2; i++; continue; } out[k++]=palLen; s=k-2; e=s-palLen; bool b=true; for (j=s; j>e; j--) { d=j-e-1; if (out[j]==d) { palLen=d; b=false; break; } out[k++]=min(d,out[j]); } if (b) { palLen=1; i++; } } out[k++]=palLen; lLen=k; s=lLen-2; e=s-(2*seqLen+1-lLen); for (i=s; i>e; i--) { d=i-e-1; out[k++]=min(d,out[i]); } } int main() { string s; cin >> s; vector<int> V(2*s.length()+1); fastLongestPalindromes(s.begin(),s.end(),V.begin()); int best=0; cout << "["; for (int i=0; i<V.size(); i++) { if (i>0) cout << ", "; cout << V[i]; best=max(best,V[i]); } cout << "]" << endl << "Longest palindrome has length " << best << endl; return 0; }
Sample input
opposes
Sample output
[0, 1, 0, 1, 4, 1, 0, 1, 0, 1, 0, 3, 0, 1, 0] Longest palindrome has length 4