LeetCode OJ 5 Longest Palindromic Substring

Posted 2023-01-27 Z-Pilgrim

https://leetcode.com/problems/longest-palindromic-substring/

感觉也许O(n)算法可以搞定，于是DP之，各种胡思乱想，于是各种BUG 。。。。。写死了，然后O(n*n)过掉的，搜下题解，然后发现一个打脸的事情，TM我写过博客http://blog.csdn.net/u011026968/article/details/26093495

O(n^2)的dp解法（对非ACM选手够了）

class Solution 
public:
    string longestPalindrome(string s) 
        // dp[i]:以i结尾的最长...  
        // dp[i] = dp[i-1] + 2 (if xxx) or 1
        if(!s.size()) return "";
        int dp[s.size()+1][s.size()+1];
        memset(dp, 0, sizeof(dp));
        int ans = 0, pos = 0;
        
        // for (int i = 0; i<s.size(); i++) 
        //     dp[i][i] = 1;
        // 
        // for(int len = 1; len <= s.size(); len++) 
        for(int i = s.size()-1; i >=0; i--) 
            for(int j=i;j< s.size(); j++) 
                if(j == i) 
                    dp[i][j] = 1;
                 else 
                    dp[i][j] = 0;
                    
                    if(s[i] == s[ j ]) 
                        if(j-i == 1)   // i+1 > j-1  2>j-i
                            dp[i][j] = 1;
                         else 
                            dp[i][j] = dp[i+1][j-1];   //max(dp[i][j], dp[i+1][j-1]+2);
                        
                        
                    
                
                
                if(dp[i][j] && j-i+1 > ans) 
                    ans = j-i+1;
                    pos = i;
                
            

        

        return s.substr(pos, ans);
    
;

就是这个O(n)算法啊，TMD  真是一朝不做ACM，各种算法全忘光

O(n)算法如下，没的说，裸地暴力

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cstring>;

using namespace std;

class Solution 
public:
    string longestPalindrome(string s) 
        int st=0,mx=1;
        int dp[2005];
        if(s.size() == 0)return "";
        dp[0]=1;
        for(int i=0;i<s.size();i++)
            if(i+1<s.size())
                int cnt=0,flag=0;
                if(s[i] == s[i+1])
                    cnt=2;
                    flag=1;
                    for(int j=i-1;j>=0 && j+cnt+1<s.size();j--)
                        if(s[j] == s[j+cnt+1])cnt+=2;
                        else break;
                    
                    dp[i]=cnt;
                    cnt=1;
                    for(int j=i-1;j>=0 && j+cnt+1<s.size();j--)
                        if(s[j] == s[j+cnt+1])cnt+=2;
                        else break;
                        //cout << i << "  " << j << "*" << cnt << endl;
                    
                    if(cnt > dp[i])dp[i]=cnt;
                    //cout << i << "  " << j << "*" << cnt << endl;
                else
                    cnt=1;
                    for(int j=i-1;j>=0 && j+cnt+1<s.size();j--)
                        if(s[j] == s[j+cnt+1])cnt+=2;
                        else break;
                    
                    dp[i]=cnt;
                
                if(dp[i]>mx)
                    /*if(flag)
                        st = i-dp[i]/2+1;
                    else
                        st = i-dp[i]/2;
                    */
                     st = i-(dp[i]+1)/2+1;
                    mx = dp[i];
                
            else
                dp[i]=1;
            
            //mx = max(mx, dp[i]);

        
        //cout << st << "  " << mx << endl;
        return s.substr(st,mx);
    
;

int main()
    freopen("5in.txt","r",stdin);
    string s;
    Solution so;
    while(cin >> s)
        cout << so.longestPalindrome(s) << endl;
    
    return 0;