<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>> results;

    vector<vector<string>> partition(const string &s) {
        vector<string> base;

        for (char i: s) {
            base.push_back(string(1, i));
        }

        recursion(base, {});
        return results;
    }

    void recursion(vector<string> remaining, vector<string> path) {
        if (remaining.empty()) {
            results.push_back(path);
            for (const string &i: path) { cout << i << " "; }
            cout << endl;
            return;
        }
        
        string tmp;


        for (int right = 0; right < remaining.size(); right++) {
            tmp += remaining[right];
            string tmp2 = tmp;
            reverse(tmp2.begin(), tmp2.end());
//            cout << "--" << tmp << "--" << tmp2 << endl;


            if (tmp == tmp2) {
                vector<string> new_path = path; // 创建新的 path 变量
                vector<string> tmp_vec(remaining.begin() + right + 1, remaining.end());
                new_path.push_back(tmp);
                recursion(tmp_vec, new_path);
            }
        }
    }
};
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