Comment on finding the longest palindrome in a string

Class Solution {Public:string Longestpalindrome (string s) {string post;    Post.resize (S.size () *2+2);    Post[0] = ' & ';      for (int i = 0;i<s.size (); i++) {post[2*i + 1] = s[i];    Post[2*i + 2] = ' # ';    } post[s.size () * * = ' # ';    Post[s.size () *2+1]= ' $ ';    cout << "Post:" << post.length () << Endl;    Vector<int> Rad;    Rad.resize (Post.size ());    Rad[0] = 0;    int Max_rad = 0;    int center = 1;    int j = 0;       for (int i = 1; i < Post.length ()-1; i) {while (post[i + j] = Post[i-j]) {j + +;       } Rad[i] = j-1;       cout << i << "," << rad[i];        if (rad[i]> max_rad) {Center = i;      Max_rad = Rad[i];      } int k;        for (k = 1;k < J && Rad[i-k]! = rad[i]-k;k++) {Rad[i+k] = min (Rad[i-k], Rad[i]-K);       cout << i+k << "," << rad[i+k];      } i = i + K;    j = Max (0, j-k);    } string Res; if (center%2 == 1) {res = S.SUBSTR (CENTER/2-MAX_RAD/2, max_rad/2*2+1);    } else {res = S.SUBSTR (center + 1)/2-(max_rad+1)/2, (max_rad+1)/2*2);  } return res; }};

The solution of the problem is actually the manacher algorithm, which is said to be the complexity of O (n), which is characterized by the use of old information in the new palindrome decision.

Problem solving:

1 duality problem, there are actually two dual forms, one is character-centric, and the other is centered on the middle position of two adjacent characters. By inserting # in the middle of the character to turn the problem into the first duality, note that the last character must be inserted in # so that it is guaranteed to be correct, because if it is #b#b, this pattern, in fact, is not equivalent to the original duality problem.

2 using the historical information of duality

The historical information of duality illustrates the precondition of the duality problem to be judged. When Rad[i-k]!=rad[i]-K, indicating I position the longest palindrome, including the i+k position of the longest palindrome, the specific analysis can refer to the longest manacher algorithm, the i+k position of the longest palindrome directly without calculation. When Rad[i-k]=rad[i]-k, the i+k position of the palindrome length at least accumulate rad[i]-k size, so you can directly from the size of the rad[i]-k to start probing.

Essentially, the nature of the palindrome duality is used in the iterative process.

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Comment on finding the longest palindrome in a string