manacher算法的解释见
这里。
//求字符串s中最大回文的长度,要求字符串s不包含字符‘#’ int manacher(const string &s) { if (s.size() <= 1) return s.size(); //往s每个字符之间以及s的首尾都插入‘#’ string str(s.size() * 2 + 1, '#'); for (int i = 0, j = 1; i<s.size(); ++i, j += 2) str[j] = s[i]; int maxr = 2, id = 1, size = str.size();//maxr是当前最大回文的半径 vector<int> p(size, 1); p[1] = 2; for (int i = 2; i <= size-3; ++i) { int maxright = p[id] + id - 1; if (i>maxright) { while (i - p[i] >= 0 && i + p[i]<size && str[i + p[i]] == str[i - p[i]]) ++p[i]; } else { int idleft = id - p[id] + 1; int k = i - id, j = id - k, tmp = j - p[j] + 1;//i和j关于id对称 if (tmp>idleft) p[i] = p[j]; else if (tmp<idleft) p[i] = p[id] - k; else { p[i] = p[j]; while (i - p[i] >= 0 && i + p[i]<size && str[i + p[i]] == str[i - p[i]]) ++p[i]; } } if (p[i] + i>p[id] + id) id = i; maxr = max(maxr, p[i]); } return maxr - 1; }
原文地址:http://blog.csdn.net/bupt8846/article/details/46374995