Shortest Palindrome

Shortest Palindrome

Total Accepted: 13465 Total Submissions: 74427 Difficulty: Hard

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".

 

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 (M) Longest Palindromic Substring (E) Implement strStr()

 

先用manacher求出以第一个字符开头的最长回文的长度，然后把剩下的字符逆序放在原始串的前面

class Solution {
public:
    void preProcess(string &s,int sSize,string& copedStr){
        for(int i=0;i<sSize;i++){
            copedStr.push_back('*');
            copedStr.push_back(s[i]);
        }
        copedStr.push_back('*');
    }
    int longestPalindromeLengthStartWithFirstChar(string s) {
        int sSize = s.size();
        if(sSize<=1){
            return sSize;
        }
        string str;
        preProcess(s,sSize,str);
        int strSize = 2*sSize+1;
        vector<int> parr(strSize,1);
        int mid = 0;
        int left =-1,right = 1;
        int pl = -1, pr = 1;
        int radius = 0;
        for(;right<strSize;right++){
            radius = 0;
            if(right>=pr){
                while((right-radius>=0 && right+radius<strSize) && str[right-radius]==str[right+radius]){
                    radius++;
                }
                parr[right] = radius;
            }else{
                left = mid - (right-mid);
                if(left - parr[left] +1 < pl+1){
                    parr[right] = left - pl;
                }else if(left - parr[left] +1 > pl+1){
                    parr[right] = parr[left];
                }else{
                    radius = parr[left]-1;
                    while((right-radius>=0 && right+radius<strSize) && str[right-radius]==str[right+radius]){
                        radius++;
                    }
                    parr[right] = radius;
                }
            }
            radius = parr[right] ;
            if(right + radius >= pr){
                pr = right + radius ;
                pl = right - radius ;
                mid = right;
            }
        }
        int maxLen = 1;
        for(int i=1;i<strSize;i++){
            if(i-parr[i]+1<=1){
                int tmpRadius = i%2==0 ? parr[i]-1:parr[i];
                if(tmpRadius>maxLen ){
                    maxLen = parr[i]-1;
                }
            }
        }
        return maxLen;
    }
    string shortestPalindrome(string s) {
        int j = longestPalindromeLengthStartWithFirstChar(s);
        string tmpStr;
        cout<<j<<endl;
        for(int i=s.size()-1;i>=j;i--){
            tmpStr.push_back(s[i]);
        }
        return tmpStr+s;
    }
};