[LeetCode-131] Palindrome Partitioning

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"]
  ]

 

历遍的说……

 

class Solution {
    public:
        vector<vector<string> > result;

        void search_palin(const string& s, int start, vector<string> parti,
                const vector<vector<bool> >& is_palin)
        {
            if (start == s.length()) {
                result.push_back(parti);
                return;
            }
            for (int i = start; i < s.length(); ++i) {                              
                if (is_palin.at(start).at(i)) {
                    parti.push_back(s.substr(start, i - start + 1));                
                    search_palin(s, i + 1, parti, is_palin);                        
                    parti.pop_back();                                               
                }
            }
        }
        
        vector<vector<string>> partition(string s) {                                
            // Start typing your C/C++ solution below                               
            // DO NOT write int main() function                                     
    
            // if char i to char j palin                                            
            int slen = s.length();                                                  
            vector<vector<bool> > is_palin(slen, vector<bool>(slen, false));        
            for (int j = 0; j < slen; ++j) {                                        
                for (int i = 0; i <= j; ++i) {                                      
                    if (i == j) {                                                   
                        is_palin.at(i).at(j) = true;                                
                    } else if (i + 1 == j) {                                        
                        is_palin.at(i).at(j) = s[i] == s[j];                        
                    } else {
                        is_palin.at(i).at(j) = is_palin.at(i + 1).at(j - 1) && s[i] == s[j];
                    }
                }
            }

            vector<string> parti;
            result.clear();
            parti.clear();
            search_palin(s, 0, parti, is_palin);
            return result;
        }
};