Leetcode242: Shortest Palindrome

Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation.

For example:

Given "aacecaaa", return "aaacecaaa".

Given "abcd", return "dcbabcd".

class Solution {public:    bool isPalindrome(string s, int start, int end)    {        while(start<end)        {            if(s[start] != s[end])  return false;            start++;    end--;        }        return true;    }    string shortestPalindrome(string s) {        int pos = s.length()-1;        if(pos==0)  return s;        for(; pos>0; pos--)            if(s[pos]==s[0] && isPalindrome(s, 0, pos)) break;        string res;        for(int i = s.length()-1; i > pos; i--)            res.push_back(s[i]);        res+=s;        return res;    }};

这种方法时间复杂度O(n2)，提交超时。而求最长回文的manacher算法时间复杂度是O(n)。

class Solution {public:    int longestPalindrom(string s) {        string s1;        s1.resize(2 * s.length() + 2);        int idx = 0;        s1[idx++] = '$';        s1[idx++] = '#';        for (char a : s) {            s1[idx++] = a;            s1[idx++] = '#';        }        vector<int> p(s1.length(), 0);        int res = 0;        for (int id = 0, i = 1; i < s1.length(); ++i) {            if (i < id + p[id]) p[i] = min(p[2 * id - i], id + p[id] - i);            else p[i] = 1;            while (s1[i + p[i]] == s1[i - p[i]]) ++p[i];            if (id + p[id] < i + p[i]) id = i;            if (p[i] == i) res = max(res, i);        }        return res - 1;    }        string shortestPalindrome(string s) {        int pos = longestPalindrom(s);        string res;        for (int i = s.length() - 1; i >= pos; --i) res.push_back(s[i]);        return res + s;    }};