leetcode[131]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"]

  ]

class Solution {

public:

void judge(string s,bool ** &T)

{

    if(s.empty())return;

    for(int i=0;i<s.length();i++)

    {

        T[i][i]=true;

        int left=i-1,right=i;

        while(left>=0&&right<s.length()&&s[left]==s[right])

        {

            T[left--][right++]=true;

        }

        left=i-1,right=i+1;

        while(left>=0&&right<s.length()&&s[left]==s[right])

        {

            T[left--][right++]=true;

        }

    }

}

void solve(vector<vector<string>> &res,vector<string> tmp,string s,bool ** &T,int pos)

{

    if(pos==s.length())

    {

        res.push_back(tmp);

        return;

    }

    for(int i=pos;i<s.length();i++)

    {

        if(!(s.substr(pos,i-pos+1)).empty()&&T[pos][i])

        {

            tmp.push_back(s.substr(pos,i-pos+1));

            solve(res,tmp,s,T,i+1);

            tmp.pop_back();

        }

    }

}

    vector<vector<string>> partition(string s) {

        vector<vector<string>> res;

        if(s.empty())return res;

        vector<string> tmp;

        int n=s.length();

        bool **T=new bool*[n];

        for(int i=0;i<n;i++)

        {

            T[i]=new bool[n];

            for(int j=0;j<n;j++)

            T[i][j]=false;

        }

        judge(s,T);

        solve(res,tmp,s,T,0);

        return res;

    }

/**

bool valid(string s)

{

    int i=0;

    int j=s.size()-1;

    while(i<j)

    {

        if(s[i++]==s[j--])continue;

        else return false;

    }

    return true;

}

void find(vector<vector<string>> &res, vector<string> tmp,string s)

{

    if(s.empty())

    {

        if(!tmp.empty())res.push_back(tmp);

        return;

    }

    for(int i=1;i<=s.size();i++)

    {

        string str=s.substr(0,i);

        if(!str.empty()&&valid(str))

        {

            tmp.push_back(str);

            find(res,tmp,s.substr(i));

            tmp.pop_back();

        }

    }

}

    vector<vector<string>> partition(string s) {

        vector<vector<string>> res;

        if(s.empty())return res;

        vector<string> tmp;

        find(res,tmp,s);

        return res;

    }

*/

};