题目描述

给定一个字符串 s，找到 s 中最长的回文子串。你可以假设 s 的最大长度为 1000。

来源，leetcode 每日一题 5. 最长回文子串

例如：

输入: "babad"
输出: "bab"
注意: "aba" 也是一个有效答案。

解题思路

1. If “aba” is a palindrome, is “xabax” and palindrome? Similarly is “xabay” a palindrome?
2. 中心向外扩展。

   manacher解法

   https://leetcode-cn.com/problems/longest-palindromic-substring/solution/zui-chang-hui-wen-zi-chuan-by-leetcode-solution/

代码

class Solution {
public:
    int expand(const string& s, int left, int right) {
        while (left >= 0 && right < s.size() && s[left] == s[right]) {
            --left;
            ++right;
        }
        return (right - left - 2) / 2;
    }
    string longestPalindrome(string s) {
        int start = 0, end = -1;
        string t = "#";
        for (char c: s) {
            t += c;
            t += '#';
        }
        t += '#';
        s = t;
        vector<int> arm_len;
        int right = -1, j = -1;
        for (int i = 0; i < s.size(); ++i) {
            int cur_arm_len;
            if (right >= i) {
                int i_sym = j * 2 - i;
                int min_arm_len = min(arm_len[i_sym], right - i);
                cur_arm_len = expand(s, i - min_arm_len, i + min_arm_len);
            } else {
                cur_arm_len = expand(s, i, i);
            }
            arm_len.push_back(cur_arm_len);
            if (i + cur_arm_len > right) {
                j = i;
                right = i + cur_arm_len;
            }
            if (cur_arm_len * 2 + 1 > end - start) {
                start = i - cur_arm_len;
                end = i + cur_arm_len;
            }
        }
        string ans;
        for (int i = start; i <= end; ++i) {
            if (s[i] != '#') {
                ans += s[i];
            }
        }
        return ans;
    }
};