[LeetCode] Shortest Palindrome

The idea is to find the longest palindromic substring of s that begins with s[0]. Then take the remaining susbtring, reverse it and append it to the beginning of s.

For example, given s = "aacecaaa", the longest palindromic substring beginning with s[0] = 'a' is "aacecaa" and the remaining substring is "a". Reverse it and append it to the beginning of s gives "aaacecaaa".

For s = "abcd", the longest palindromic substring beginning with s[0] = 'a' is "a" and the remaining substring is "bcd". Reverse it and append it to the beginning of s gives "dcbabcd".

The most difficult part is to implement the Manacher's algorithm to find the longest palindromic substring starting with s[0]. Please refer to this nice article if you want to know how it works.

The code is as follows.

class Solution {
public:
    string shortestPalindrome(string s) {
        string t = process(s);
        int n = t.length(), center = 0, right = 0;
        int* palin = new int[n];
        for (int i = 1; i < n - 1; i++) {
            int i_mirror = 2 * center - i;
            palin[i] = (right > i) ? min(palin[i_mirror], right - i) : 0;
            while (t[i + palin[i] + 1] == t[i - palin[i] - 1])
                palin[i]++;
            if (i + palin[i] > right) {
                center = i;
                right = i + palin[i];
            }
        }
        int pos;
        for (int i = n - 2; i; i--) {
            if (i - palin[i] == 1) {
                pos = palin[i];
                break;
            }
        }
        string tail = s.substr(pos); 
        reverse(tail.begin(), tail.end());
        return tail + s;
    }
private:
    string process(string& s) {
        int n = s.length();
        string t(2 * n + 3, '#');
        t[0] = '$'; t[2 * n + 2] = '%';
        for (int i = 0; i < n; i++)
            t[2 * (i + 1)] = s[i];
        return t;
    }
};

Note that this part of the code is just to find the ending position of the longest palindromic substring begining with s[0].

int pos;
for (int i = n - 2; i; i--) {
    if (i - palin[i] == 1) {
        pos = palin[i];
        break;
    } 
}