<p>Given a string <code>s</code>, partition <code>s</code> such that every <span data-keyword="substring-nonempty">substring</span> of the partition is a <span data-keyword="palindrome-string"><strong>palindrome</strong></span>. Return <em>all possible palindrome partitioning of </em><code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aab"
<strong>Output:</strong> [["a","a","b"],["aa","b"]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "a"
<strong>Output:</strong> [["a"]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 16</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution {
public:
    bool is_palindrom(int first, int last, string& s) {
        if (first == last) { return true; }
        while (first < last) {
            if (s[first] != s[last]) { return false; }
            ++first;--last;
        }
        return true;
    }



    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        int n = s.size();
        if (n == 1) {
            ans.push_back(vector<string>{s});
            return ans;
        }
        vector<vector<string>> dop;
        int first = 0;
        int second = 0;
        while (second < n) {
            if (is_palindrom(first, second, s) == true) {
                string g;
                for (int i = 0;i <= second;++i) {
                    g.push_back(s[i]);
                }
                if (second == n - 1) {
                    ans.push_back(vector<string>{g});
                }else{
                    dop.push_back(vector<string>{g});
                }
            }
            ++second;

        }

        int start = 0;
        while (start < dop.size()) {
            int t = dop.size();
            for (start;start < t;++start) {
                first = 0;
                for (int w = 0;w < dop[start].size();++w) {
                    first += dop[start][w].size();
                }
                
                second = first;
                while (second < n) {
                    if (is_palindrom(first, second, s) == true) {
                        string g;
                        for (int i = first;i <= second;++i) {
                            g.push_back(s[i]);
                        }
                        vector<string> d = dop[start];
                        d.push_back(g);
                        if (second == n - 1) {
                            ans.push_back(d);
                        }
                        else {
                            dop.push_back(d);
                        }
                    }
                    ++second;

                }
            }
        }
        return ans;
    }
};
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