<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:
    vector<vector<string>> partition(string s) {
        cur.emplace_back(1,s[0]);
        dfs(s,1);
        return res;
    }
    void dfs(string& s, int i){
        if(cur.size() > 1 and !isPalindrome(*(cur.end()-2))) return;
        if(i == s.size()){ 
            if(isPalindrome(cur.back())) res.push_back(cur); 
            return;
        }
        string t = *(cur.end() - 1);
        cur.emplace_back(1,s[i]);
        dfs(s,i+1);
        cur.erase(cur.end()-1);
        *(cur.end()-1) += s[i];
        dfs(s,i+1);
        *(cur.end()-1) = t;
    }
private:
    bool isPalindrome(string& s) {
        int l = 0;
        int r = s.size()-1;
        while(l<r) if(s[l++] != s[r--]) return false;
        return true;
    };
    vector<vector<string>> res;
    vector<string> cur;
};
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