DP + Backtrack . Beats 59%


  • 0
    A

    Pos denotes the index from where we should consider to form a palindrome

     class Solution {
        public:
            void go(string s,int pos,vector<string> &x,vector<vector<string>> &ans,vector<vector<bool>> &dp)
            {
                if(pos>s.size()) return;
                if(pos==s.size())
                {
                    if(x.size()>0)
                        ans.push_back(x);
                    return;
                }
                for(int i=pos;i<s.size();i++)
                {
                    if(dp[pos][i])
                    {
                        string t= s.substr(pos,i+1-pos);
                        x.push_back(t);
                        go(s,i+1,x,ans,dp);
                        x.pop_back();
                    }
                    
                }
            }
            vector<vector<string>> partition(string s) {
                vector<vector<bool>> dp(s.size(),vector<bool>(s.size(),false));
                vector<vector<string>> ans;
                int n = s.size();
                if(!n) return ans;
                for(int i=0;i<n;i++) dp[i][i]=true;
                for(int i=0;i<n-1;i++) if(s[i]==s[i+1]) dp[i][i+1]=true;
                for(int l=3;l<=n;l++)
                {
                    for(int i=0;i<=n-l;i++)
                    {
                        int j= i+l-1;
                        if(dp[i+1][j-1]== true && s[i]==s[j]) dp[i][j]=true;
                    }
                }
                vector<string> x;
                go(s,0,x,ans,dp);
                return ans;
            }
        };

Log in to reply
 

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.