5. 最长回文子串

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。

示例 1：

输入: "babad"
输出: "bab"
注意: "aba" 也是一个有效答案。
示例 2：

输入: "cbbd"
输出: "bb"

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/longest-palindromic-substring

s0：暴力解法

列举所有子串，判断是否为回文

class Solution {
    bool isPalindromic(string str)
    {
        for(int i = 0; i < str.size() / 2; i++)
        {
            if(str[i] != str[str.size() - 1 - i])
            {
                return false;
            }
        }
        return true;
    }
public:
    string longestPalindrome(string s) {
        string ans = "";
        int max = 0;
        int len = s.size();
        for(int i = 0; i < len; i++)
        {
            for(int j = 1; i + j <= len; j++)
            {
                string tmp_str = s.substr(i, j);
                if(isPalindromic(tmp_str) && tmp_str.size() > max)
                {    
                    max = tmp_str.size();
                    ans = tmp_str;
                }
            }
        }
        return ans;
    }
};

s1：最长公共子串

1、dp找最长公共字串

#include <bits/stdc++.h>

using namespace std;

string longestPubSubStr(string s1, string s2)
{
    string ret;
    int m = s1.size();
    int n = s2.size();
    int dp[m][n] = {0};
    fill(dp[0], dp[0] + m * n, 0);
    
    for(int i = 0; i < m; i++)
    {
        if(s1[i] == s2[0])
        {
            dp[i][0] = 1;
        }
    }
    for(int j = 0; j < n; j++)
    {
         if(s1[0] == s2[j])
        {
            dp[0][j] = 1;
        }
    }
    int maxLen = 0;
    int maxLenEndCharIndex = 0;
    for(int i = 1; i < m; i++)
    {
        for(int j = 1; j < n; j++)
        {
            if(s1[i] == s2[j])
            {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                if(dp[i][j] > maxLen)
                {
                    maxLen = dp[i][j];
                    maxLenEndCharIndex = i;
                }
            }
        }
    }
    
    
    ret = s1.substr(maxLenEndCharIndex - maxLen + 1, maxLen);
    
    return ret;
}

int main()
{
    string s1 = "abac";
    string s2 = "caba";
    
    cout << longestPubSubStr(s1, s2) << endl; 
    
    return 0;
}

这里的空间复杂度可以优化

#include <bits/stdc++.h>

using namespace std;

int longestPubSubStr(string s1, string s2)
{
    int max_len = 0;
    int n = s1.size();
    int dp[n] = {0};
    
    // i 是列数， j 是行数，按列更新 
    for(int i = 0; i < n; i++)
    {
        for(int j = n - 1; j >= 0; j--)
        {
            if((i == 0 || j == 0) && s1[i] == s2[j])
            {
                dp[j] = 1;
                if(dp[j] > max_len) max_len = dp[j];
            }
            else if(s1[i] == s2[j])
            {
                dp[j] = dp[j - 1] + 1;
                if(dp[j] > max_len) max_len = dp[j];
            }
            else
            {
                dp[j] = 0; //这里使用相同的列需要置0 
            }
        }
    }
    
    return max_len;
}

int main()
{
    string s1 = "abacssss";
    string s2 = "cabassss";
    
    cout << longestPubSubStr(s1, s2) << endl; 
    
    return 0;
}

2、判断最长公共字串是否对应

class Solution {
    string longestPubSubStr(string s1, string s2)
    {
        string ret;
        int m = s1.size();
        int n = s2.size();
        cout << m << n << endl;
        int dp[m][n] = {0};
        fill(dp[0], dp[0] + m * n, 0);
        int maxLen = 0;
        int maxLenEndCharIndex = 0;
        
        for(int i = 0; i < m; i++)
        {
            if(s1[i] == s2[0])
            {
                dp[i][0] = 1;
                if(dp[i][0] > maxLen)
                {
                    maxLen = dp[i][0];
                    maxLenEndCharIndex = i;
                }
            }
        }
        for(int j = 0; j < n; j++)
        {
            if(s1[0] == s2[j])
            {
                dp[0][j] = 1;
            }
        }
        
        for(int i = 1; i < m; i++)
        {
            for(int j = 1; j < n; j++)
            {
                if(s1[i] == s2[j])
                {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    if(dp[i][j] > maxLen)
                    {
                        int beforeI = m - 1 - i;
                        if(beforeI + dp[i][j] - 1 == j)
                        {
                            maxLen = dp[i][j];
                            maxLenEndCharIndex = i;
                        }
                    }
                }
            }
        }

        ret = s1.substr(maxLenEndCharIndex - maxLen + 1, maxLen);
        
        return ret;
    }
public:
    string longestPalindrome(string s) {
        if(s.size() <= 1) return s;
        string s1 = s;
        reverse(s.begin(), s.end());
        return longestPubSubStr(s, s1);
    }
};