LeetCode:Palindrome Partitioning(DP TLE)

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

动规还TLE 我也是醉了!

class Solution {
public:
    vector<vector<string>> partition(string s) {
        //dp解法
        /*用来记录s[i]到s[j]是否为palindrome
        dp[i][j]   1、s[i]==s[j]&&j-i<2
        2、s[i]==s[j]&&s[i+1]到s[j-1]为palindrome 
        所以事先要知道dp[i+1][j-1] i是s.length-- j从i++
        */
        const int n=s.size();
        
        bool dp[n][n];
        fill_n(&dp[0][0],n*n,false);
        for(int i=s.size()-1;i>=0;i--)
            for(int j=i;j<n;++j)
                if(s[i]==s[j]&&(j-i<2)||dp[i+1][j-1])
                    dp[i][j]=true;
            
                    
       //开始根据dp[i][j]截取字符串 获得结果
       
       vector<vector<string>> sub_palins[n]; 
       
       for(int i=n-1;i>=0;i--)
        for(int j=i;j<n;++j)
            if(dp[i][j])
            {
                const string palindrome=s.substr(i,j-i+1);
                if(j+1<n)
                    for(auto v:sub_palins[j+1])
                        {
                            v.insert(v.begin(),palindrome);
                            sub_palins[i].push_back(v);
                        }
                else{
                        sub_palins[i].push_back(vector<string>{palindrome});
                }
            }
            
            return sub_palins[0];
    }
};
原文地址:https://www.cnblogs.com/xiaoying1245970347/p/4748170.html