Sunday, February 15, 2015

LeetCode [5] 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.
Example 1:
Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Example 2:
Input: "cbbd"
Output: "bb"
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class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.size();
        string ret;
        vector<vector<int>> dp(n, vector<int>(n, 0));
        for(int i=n-1; i>=0; --i){
            for(int j=i; j<n; ++j){
                if(i!=j){
                    if(s[i]==s[j]){
                        if(i+1==j) dp[i][j] = 2;
                        else if(dp[i+1][j-1]>0) dp[i][j]=dp[i+1][j-1]+2;
                    }
                }else{
                    dp[i][j]=1;
                }
                
                if(ret.size()<dp[i][j]) ret = s.substr(i, dp[i][j]);
            }
        }
        return ret;
    }
};

class Solution {
public:
    string longestPalindrome(string s) {
        int start, len = 0, n = s.size();
        bool dp[n][n];
        memset(dp, false, n*n*sizeof(bool));
        const char *d = s.data();
        
        for(int i=0; i<n; ++i){
            dp[i][i] = true;
            if(len<1){start = i; len = 1;}
        }
        
        for(int i=0; i<n-1; ++i){
            if(d[i]==d[i+1]){
                dp[i][i+1] = true;
                if(len<2){start = i; len = 2;}
            }
        }
        
        for(int i=n-3; i>=0; --i){
            for(int j=i+2; j<n; ++j){
                if(d[i]==d[j] && dp[i+1][j-1]){
                    dp[i][j] = true;
                    if(len<j-i+1){len = j-i+1; start = i;}
                }
            }
        }
        
        return s.substr(start, len);
    }
};

class Solution {
public:
    string longestPalindrome(string s) {
        bool dp[1000][1000] = {false};
        int n = s.size();
        string ret;
        for(int i=n-1; i>=0; --i){
            for(int j=i; j<n; ++j){
                if(s[i]==s[j]){
                    if(i+1>j-1 || dp[i+1][j-1]){
                        dp[i][j] = true;
                        if(j-i+1>(int)ret.size()){
                            ret = s.substr(i, j-i+1);
                        }
                    }
                }
            }
        }
        return ret;
    }
};

class Solution {
    int expand(int &l, int &r, string s){
        int len = -1;
        int n = s.size();
        while(l>=0 && r<n && s[l]==s[r]){
            len = r-l+1;
            l--;
            r++;
        }
        l++;
        r--;
        return len;
    }
public:
    string longestPalindrome(string s) {
        int n = s.size();
        if(n<=1) return s;
        int start = 0, len = 1;
        for(int i=0; i<n; ++i){
            int l = i;
            int r = i;
            int len1 = expand(l, r, s);
            if(len1>len){
                start = l;
                len = len1;
            }
            l = i;
            r = i+1;
            len1 = expand(l, r, s);
            if(len1>len){
                start = l;
                len = len1;
            }            
        }
        return s.substr(start, len);
    }
};

class Solution {
public:
    string expanding(string s){
        string T;
        for(int i=0; i<s.size(); ++i){
            T += '#';
            T += s[i];
        }
        T += "#";
        return T;
    }

    string longestPalindrome(string s) {
        string T = expanding(s);
        int n = T.size();
        int dp[n];
        memset(dp, 0, n*sizeof(int));

        int C = 0, R = 0, i = 1, maxLen = 0, start = 0;
        for(int i=1; i<n; ++i){
            int i_mirror = C*2-i;
            dp[i] = (R>i)?min(R-i, dp[i_mirror]):0;
            while(i-dp[i]-1>=0 && i+dp[i]+1<n && T[i-dp[i]-1]==T[i+dp[i]+1]){
                dp[i]++;
            }
            if(i+dp[i]>R){
                C = i;
                R = i+dp[i];
            }
            if(dp[i]>maxLen){
                maxLen = dp[i];
                start = (i-maxLen)/2;
            }
       }
        return s.substr(start, maxLen);
    }
};
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