「回溯法」Palindrome Partitioning

/*************************************************************************
	> File Name: PalindromePartitioning.cpp
	> Author: Shaojie Kang
	> Mail: [email protected] 
	> Created Time: 2015年09月16日 星期三 16時46分38秒 
    > Problem:
        Given a string s, partition s such that every substring of the partition is a palindrome.

        Return all possible palindrome partitioning of s.

    Have you met this question in a real interview? Yes
    Example
    given s = "aab",
    Return

  [
      ["aa","b"],
      ["a","a","b"]
  ]
 ************************************************************************/

#include<iostream>
#include<vector>
#include<string>
using namespace std;

class Solution 
{
public:
    bool isPalindrome(const string &str)
    {
        if(str.empty()) return true;
        int size = str.size();
        for(int i = 0, j = size-1; i < j; ++i, --j)
        {
            if(str[i] != str[j]) return false;
        }
        return true;
    }

    vector<vector<string> > partition(string s)
    {
        vector<vector<string> >result;
        vector<string> oneResult;
        partitionCore(s, 0, result, oneResult);

        return result;
    }

    void partitionCore(string &s, int start, vector<vector<string> > &result, vector<string> oneResult)
    {
        int size = s.size();
        if(start == size)
        {
            result.push_back(oneResult);
            return;
        }
        for(int i = start; i < size; ++i)
        {
            string strTemp(s.substr(start, i-start+1));
            if(isPalindrome(strTemp))
            {
                oneResult.push_back(strTemp);
                partitionCore(s, i+1, result, oneResult);
                oneResult.pop_back();
            }
        }
    }
};

void print(const vector<vector<string> > &result)
{
    for(int i = 0; i < result.size(); ++i)
    {
        for(int j = 0; j < result[i].size(); ++j)
        {
            cout<<result[i][j]<<" ";
        }
        cout<<endl;
    }
}

int main()
{
    string str("aab");
    Solution sol;
    vector<vector<string> > result = sol.partition(str);
    print(result);

    return 0;
}