Palindrome Partitioning

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"]
  ]

填写dp矩阵的时候坐标弄反了浪费了好多时间……

class Solution {
public:
    vector<vector <int> >dp;
    vector<vector<string>> re;
    int len;
    vector<vector<string>> partition(string s) {
        
        string ss ="";
        if(s.size() == 0)return re;
        if(s.size()==1)
        {
            vector<string> tp ;
            tp.push_back(s);
            re.push_back(tp);
            return re;
        }
        ss = ss+s[0];
        for(int i = 1 ; i  < s.length();i++)
        {
            ss = ss + "#";
            ss = ss + s[i];
        }
        len = ss.size();
        
        
        for(int i = 0 ; i < len ;i++)
        {
            vector<int> tp(len,0);
            dp.push_back(tp);
        }
        for(int i  = 0  ; i < len ; i++)
        dp[i][i] = 1;
        
        for(int i = 0  ; i < len ;i++)
        {
            int j = 1;
            while(i-j>=0 && i+j<len && ss[i-j] == ss[i+j])
            {
                dp[i-j][i+j] = 1;
                j++;
            }
        }
        vector<string> temp;
        fill(0,temp,s);
        return re;
    }
    void fill(int n, vector<string> now,string s)
    {
        if(n >= len)
        {
            re.push_back(now);
            return;
        }
        for(int i = n ; i < len ;i++)
        {
            if(dp[n][i] == 1)
            {
                if(n%2 == 0)
                    {
                        string sub =s.substr(n/2,i/2-n/2+1);
                        if(sub == "")continue;
                        now.push_back(sub);
                    }
                    else
                    {
                        string sub =s.substr((n+1)/2,(i+1)/2-(n+1)/2);
                        if(sub == "")continue;
                        now.push_back(sub);
                    }
                    fill(i+2,now,s);
                    now.pop_back();
            }
        }
    }
};