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53 lines
1.1 KiB
C++
53 lines
1.1 KiB
C++
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#include <iostream>
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#include <string>
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using std::cin;
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using std::cout;
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const char endl = '\n';
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const int N = 1e7 + 5;
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int p[N << 1], mid, r, ans;
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std::string s1, s2;
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int main() {
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std::ios::sync_with_stdio(false);
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cin.tie(nullptr);
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cin >> s1;
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// 使字符串的长度变为 (2n + 1),方便处理
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s2.push_back('^');
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for (const char &c : s1) {
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s2.push_back('#');
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s2.push_back(c);
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}
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s2 += "#$";
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for (int i = 1; i < s2.size(); i++) {
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// r 代表以 mid 为中心的最长回文子串的右边界
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p[i] = i < r
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// mid * 2 - i 为 i 关于 mid 的对称点
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? std::min(p[mid * 2 - i], r - i)
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// 超过边界就不是回文串了
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: 1;
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// 扩展回文串长度
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while (s2[i - p[i]] == s2[i + p[i]]) p[i]++;
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// 扩展右边界
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if (r < i + p[i]) {
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r = i + p[i];
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mid = i;
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}
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}
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for (int i = 0; i < s2.size(); i++) {
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ans = std::max(ans, p[i] - 1); // p[i] - 1 即为以 i 为中心的最长的回文子串长度
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}
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cout << ans << endl;
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return 0;
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}
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