【C++】最长回文子串/动态规划

ACM

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;
char S[maxn];
int dp[maxn][maxn];

int main()
{
    gets(S);
    int len = strlen(S), ans = 1;
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < len; i++)
    {
        dp[i][i] = 1;
        if (i < len - 1)
        {
            if (S[i] == S[i + 1])
            {
                dp[i][i + 1] = 1;
                ans = 2;
            }
        }
    }
    // 状态转移方程
    for (int L = 3; L <= len; L++)
    {
        for (int i = 0; i + L - 1 < len; i++)
        {
            int j = i + L - 1;
            if (S[i] == S[j] && dp[i + 1][j - 1] == 1)
            {
                dp[i][j] = 1;
                ans = L;
            }
        }
    }
    cout << ans;
    system("pause");
}

核心代码

#include <bits/stdc++.h>
using namespace std;

class Solution
{
public:
    int getLongestPalindrome(string A, int n)
    {
        int maxR = 1;
        // 创建dp数组
        vector<vector<int>> dp;
        vector<int> tmp;
        tmp.insert(tmp.begin(), n, 0);
        for (int i = 0; i < n; i++)
        {
            dp.push_back(tmp);
        }
        // 边界条件
        for (int i = 0; i < n; i++)
        {
            dp[i][i] = 1;
            if (i < n - 1)
            {
                if (A[i] == A[i + 1])
                {
                    dp[i][i + 1] = 1;
                    maxR = 2;
                }
            }
        }
        // 状态转移
        for (int len = 3; len <= n; len++)
        {
            // 枚举左端点i
            for (int i = 0; i + len - 1 < n; i++)
            {
                int j = i + len - 1;
                if (A[i] == A[j] && dp[i + 1][j - 1] == 1)
                {
                    dp[i][j] = 1;
                    maxR = len;
                }
            }
        }
        return maxR;
    }
};

int main()
{
    string str;
    cin >> str;
    int n = str.length();
    Solution solution;
    cout << solution.getLongestPalindrome(str, n) << endl;
    system("pause");
}