最大长度回文子串（Manacher's algorithm）

 1 string longestPalindrome(string s)
 2 {
 3 
 4 
 5     int id, mx, i, j, len, maxlen;
 6     vector<char> cvec;
 7     vector<int> ivec;
 8     string str;
 9     int ma, k;
10 
11     id = 0;
12     mx = 0;
13     ivec.push_back(1);
14     len = s.size();
15     maxlen = 2*s.size() + 1;
16 
17     for (i = 0; i < len; ++i)
18     {
19         cvec.push_back(‘#‘);
20         cvec.push_back(s[i]);
21     }
22     cvec.push_back(‘#‘);
23 
24 
25     for (i = 1; i < maxlen; ++i)
26     {
27         if (mx > i)
28         {
29             int temp;
30             temp = ((mx - i - 1) > (ivec[2*id - i])) ? (ivec[2*id - i]) : (mx - i - 1);
31             ivec.push_back(temp);
32         }
33         else
34             ivec.push_back(1); 
35 
36         for (; (i - ivec[i] >= 0) && ((i + ivec[i]) < maxlen) && (cvec[i-ivec[i]] == cvec[i+ivec[i]]); ++ivec[i])
37             ;
38 
39         if (mx < ivec[i] + i -1 )
40         {
41             mx = ivec[i] + i - 1;
42             id = i;
43         }
44     }
45 
46     ma = 0;
47     for (i = 0; i < maxlen; ++i)
48     {
49         if (ivec[i] > ma)
50         {
51             ma = ivec[i];
52             k = i;
53         }
54     }
55 
56     for (j = 0, i = k - ivec[k] + 1; j < ivec[k] - 1; ++i, ++j)
57     {
58         ++i;
59         str.push_back(cvec[i]);
60     }
61     
62     return str;
63 }