leetcode-5最长回文子串(manacher算法)

输入：s = "babad"
输出："bab"
解释："aba" 同样是符合题意的答案。
示例 2：

输入：s = "cbbd"
输出："bb"
示例 3：

输入：s = "a"
输出："a"
示例 4：

输入：s = "ac"
输出："a"

class Solution { 
   
public:
    string longestPalindrome(string s) { 
   
        int res = 0,len = 1,l = 0,r = 0;
        for(int i = 0;i < s.size();i ++){ 
   
            len = 0;
            while(i - len >= 0 && i + len < s.size() && s[i - len] == s[i + len])len ++;
            if(len * 2 - 1 > res)res = len * 2 - 1,l = i - len + 1,r = i + len - 1;
        }
        for(int i = 0;i < s.size() - 1;i ++){ 
   
            int ll = i,rr = i + 1;
            while(ll >= 0 && rr < s.size() && s[ll] == s[rr])ll --,rr ++;
            if(rr - ll - 1 > res)res = rr - ll - 1,l = ll + 1,r = rr - 1;
        }
    
        return s.substr(l,r - l + 1);
    }
};

class Solution { 
   
public:
    string longestPalindrome(string s) { 
   
        string t;
        t.append(1,'#');
        for(int i = 0;i < s.size();i ++){ 
   
            t.append(1,s[i]);
            t.append(1,'#');
        }
        cout<<t<<endl;
        vector<int>r(t.size());
        r[0] = 1;
        int res = 0;
        for(int i = 0,center = 0;i < t.size();i ++){ 
   
            if(center + r[center] < i){ 
   
                while(i + r[i] < t.size() && i - r[i] >= 0 && t[i + r[i]] == t[i - r[i]])r[i] ++;
                r[i] --;
            }
            else{ 
   
                int m = center - (i - center);
                if(m - r[m] > center - r[center])r[i] = r[m];
                else { 
   
                    r[i] = center + r[center] - i;
                    while(i + r[i] < t.size() && i - r[i] >= 0 && t[i - r[i]] == t[i + r[i]])r[i] ++;
                    r[i] --;
                }
            }
            if(r[i] > r[res])res = i;
            if(i + r[i] >= center + r[center])center = i;
            cout<<i<<" "<<r[i]<<endl;
        }
        cout<<res<<" "<<r[res]<<endl;
        string a;
        for(int i = res - r[res];i <= res + r[res];i ++){ 
   
            if(t[i] != '#')a.append(1,t[i]);
        }
        return a;
    }
};