Leetcode Longest Palindromic Substring

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.

题目的意思：输入一个字符串S，找出其最长回文串

用动态规划求解

决策变量：dp[i][j]记录从s[i]到s[j]组成的子串是否为回文。

　　　　　　dp[i+1][j-1]     , 当s[i]与s[j]相等时

dp[i][j] =

　　　　　　false　　　　　　, 当s[i]与s[j]不相等时

单个字符是回文串，故要初始化对角线

紧邻的两个相同字符也是回文

时间复杂度O(n^2)

class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.length(), startIndex =  , maxLen = ;
        // vector<vector<bool> > dp(n,vector<bool>(n,false));
        bool dp[][] = {false};
        for(int i =  ; i < n; ++ i) dp[i][i] = true;
        for(int i = ;  i < n-; ++ i ){
            if(s[i] == s[i+]){
                dp[i][i+] = true;
                startIndex= i;
                maxLen = ;
            }
        }
        for(int len = ; len <= n; ++len){
            for(int i =  ; i < n-len+; ++ i){
                int j = i+len-;
                if(s[i] == s[j] && dp[i+][j-]){
                    dp[i][j] = true;
                    startIndex =i;
                    maxLen = len;
                }
            }
        }
        return s.substr(startIndex,maxLen);
    }
};

动态规划求解

利用Manacher 算法求解，时间复杂度为O(n)

可以参考http://www.felix021.com/blog/read.php?2040

和http://leetcode.com/2011/11/longest-palindromic-substring-part-ii.html

string preProcess(string s){
    int n = s.length();
    if( n == ) return "^$";
    string res="^";
    for(int i =  ; i < n; ++ i) res+="#"+string(,s[i]);
    res+="#$";
    return res;
}

string  longestPalindrome(string s){
    string T = preProcess(s);
    int n = T.length();
    vector<int> p(n,);
    int center = , radius = ,maxv = ;
    for(int i = ; i < n-; ++ i){
        p[i] = (radius > i) ? min(radius-i,p[*center-i]) : ;
        while(T[i++p[i]] == T[i--p[i]]) p[i]++;
        if(i+p[i] > radius){
            center = i;
            radius = i+p[i];
        }
    }
    int maxLen = , centerIndex = ;
    for(int i = ; i < n-; ++ i){
        if(p[i] > maxLen){
            maxLen = p[i];
            centerIndex = i;
        }
    }
    centerIndex = (centerIndex - -maxLen)/;
    return s.substr(centerIndex,maxLen);
}

Manacher算法