<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 palindrome(string s) {
        for (int i = 0, j = s.size() - 1; i < j; i++, j--) {
            if (s[i] != s[j]) {
                return false;
            }
        }
        return true;
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        int n = s.size();
        for (int pre_mask = 0; pre_mask < (1 << (n - 1)); pre_mask++) {
            int mask = pre_mask | (1 << (n - 1));
            bool ok = true;
            vector<string> vt;
            string t;
            for (int i = 0; i < n; i++) {
                t += s[i];
                int bit = (mask >> i) & 1;
                if (bit) {
                    if (!palindrome(t)) {
                        ok = false;
                        break;
                    }
                    vt.push_back(t);
                    t.clear();
                }
            }
            if (ok) {
                ans.push_back(vt);
            }
        }
        return ans;
    }
};
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