Palindrome Partitioning

方法：使用深度遍历的方法，时间复杂度O(2^n)

class Solution {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> result;
        vector<string> path;
        
        dfs(s, path, result, 0);
        return result;
    }
    
    void dfs(string &s, vector<string> &path, vector<vector<string>> &result, int start)
    {
        if(start == s.size())
        {
            result.push_back(path);
            return;
        }
        
        for(int i=start; i<s.size(); ++i)
        {
            if(isPalindrome(s, start, i))
            {
                path.push_back(s.substr(start, i-start +1));
                dfs(s, path, result, i+1);
                path.pop_back();
            }
        }
    }
    
    bool isPalindrome(string s, int start, int end)
    {
        while(start < end && s[start] == s[end])
        {
            ++start;
            --end;
        }
        return (start >= end);
    }
};