<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 {
private:
    bool isPalindrome(string &s) {
        for (int i = 0; i < s.size()/2; i++) {
            if (s[i] != s[s.size()-1-i])
                return false;
        }
        return true;
    }

    void partitionAux(string &s, int left, int right, vector<string> &currPartition, vector<vector<string>> &allPartitions) {
        if (left == s.size()) {
            vector<string> newPartition(currPartition.size());
            copy(currPartition.begin(), currPartition.end(), newPartition.begin());
            allPartitions.push_back(newPartition);
            return;
        }

        for (int i = left; i <= right; i++) {
            string tmp = s.substr(left, i-left+1);
            if (isPalindrome(tmp)) {
                currPartition.push_back(tmp);
                partitionAux(s, i+1, right, currPartition, allPartitions);
                currPartition.pop_back();
            }
        }
    }
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> allPartitions;
        vector<string> currPartition;
        partitionAux(s, 0, s.size()-1, currPartition, allPartitions);
        return allPartitions;
    }
};
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