<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 ispalindrom(vector<string>&vs){
        for(auto str:vs){
        int l=0,r=str.size()-1;
        while(r>=l){
            if(str[l]!=str[r])return false;
            l++;
            r--;
          }
        }
        return true;
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>>ans;
        int n=s.size();
        for(int i=0;i<(1<<(n-1));i++){
           vector<string>vs;
           string str;
           for(int bit=0;bit<n;bit++){
               str.push_back(s[bit]);
               if(1&(i>>bit)){
                vs.push_back(str);
                str="";
               }
           }
           vs.push_back(str);
           if(ispalindrom(vs)){
            ans.push_back(vs);
           }
        }
        return ans;
    }
};
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