LCS length

Recursive

int lcs( char *X, char *Y, int m, int n )

{
   if (m == 0 || n == 0)
     return 0;
   if (X[m-1] == Y[n-1])
     return 1 + lcs(X, Y, m-1, n-1);
   else
     return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
}
Dynamic Programming

int lcs( char *X, char *Y, int m, int n )

{
   int L[m+1][n+1];
   int i, j;
 
   /* Following steps build L[m+1][n+1] in bottom up fashion. Note
      that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
   for (i=0; i<=m; i++)
   {
     for (j=0; j<=n; j++)
     {
       if (i == 0 || j == 0)
         L[i][j] = 0;
 
       else if (X[i-1] == Y[j-1])
         L[i][j] = L[i-1][j-1] + 1;
 
       else
         L[i][j] = max(L[i-1][j], L[i][j-1]);
     }
   }
   
   /* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */
   return L[m][n];
}
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