Longest Palindromic Substring

Total Accepted: 82026 Total Submissions: 379898 Difficulty: Medium

Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.

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 (H) Shortest Palindrome (E) Palindrome Permutation

 

1.动态规划

class Solution {
public:
    string longestPalindrome(string s) {
        int sSize = s.size();
        if(sSize <=1){
            return s;
        }
        int start = 0;
        int maxLen = 1;
        bool help[1000][1000] = {false};
        for(int i=0;i<sSize;i++){
            help[i][i] = true;
            if(i>0 && s[i-1]==s[i]){
                help[i-1][i] = true;
                start = i-1;
                maxLen = 2;
            }
        }
        for(int len = 3;len<=sSize;len++){
            for(int i=0;i<=sSize-len;i++){
                int j = i+len-1;
                if(help[i+1][j-1] && s[i]==s[j]){
                    help[i][j] = true;
                    start = i;
                    maxLen = len;
                }
            }
        }
        return s.substr(start,maxLen);
    }
};

2.中心扩展

3.manacher

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('*');
    }
    string longestPalindrome(string s) {
        int sSize = s.size();
        if(sSize<=1){
            return s;
        }
        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 = 0, pos = 0;
        for(int i=1;i<strSize;i++){
            int tmpRadius = i%2==0 ? parr[i]-1:parr[i];
            if(tmpRadius>maxLen ){
                pos = i;
                maxLen = parr[i]-1;
            }
        }
        string res;
        for(int i=pos-maxLen;i<=pos+maxLen;i++){
            if(i%2!=0){
                res.push_back(str[i]);
            }
        }
        return res;
    }
};