最长回文子串

#include <iostream>
#include <vector>
#include <string>

using namespace std;

class Solution {
public:
    string longestPalindrome(string s) {
        unsigned int n = s.size();
        if (n < 2) {
            return s;
        }
        int maxLen = 1, begin = 0;
        /**
         * 状态: dp[i][j] : 子串s[i....j] 是否是回文串.
         * 状态转移方程: dp[i][j] = (s[i] == s[j]) 并且 dp[i+1][j-1]
         * 边界条件:
         */

        vector<vector<int>> dp(n, vector<int>(n));
        // 单个的字符是回文串
        for(int i = 0; i < n; i++) {
            dp[i][i] = true;
        }
        // 枚举子字符串串长度,通过扩大子字符串的长度寻找回文串.
        for (int L = 2; L <= n; L++) {
            // 左边界
            for (int i = 0; i < n; i++) {
                // 右边界 (j - i + 1 = L)
                int j = L + i - 1;
                // 右边界越界
                if (j >= n) {
                    break;
                }
                // 子字符串的左右两边不相等
                if (s[i] != s[j]) {
                    dp[i][j] = false;
                } else {
                    // 达到边界条件
                    if (j - i < 3) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i+1][j-1];
                    }
                }
                // 子串是回文,更新回文字符串长度和起始位置.
                if (dp[i][j] && j - i + 1 > maxLen) {
                    maxLen = j - i + 1;
                    begin = i;
                }
            }
        }
        return s.substr(begin, maxLen);
    }
};

int main() {
    Solution solution;
    string s = "bacad";
    string result = solution.longestPalindrome(s);
    cout<< result << endl;
}