<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>> result;
    vector<string> path;
    bool isPal(string s, int start, int end){
        for (int i = start, j = end; i < j; i++, j--) {
            if (s[i] != s[j]) {
                return false;
            }
        }
        return true;
    }
    void part(string s, int idx){
        if(idx == s.size()) {
            result.push_back(path);
            return;
        }
        for(int i = idx; i < s.size(); i++){
            if(isPal(s, idx, i) == false) continue;
            string tmp = s.substr(idx, i - idx + 1);
            path.push_back(tmp);
            part(s, i + 1);
            path.pop_back();
        }
        return;
    }
    vector<vector<string>> partition(string s) {
        part(s, 0);
        return result;
    }
};
88341