Palindrome Partitioning
<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>
String; Dynamic Programming; Backtracking
class Solution {
    private int n;
    private String s;
    private boolean[][] f;
    private List<String> t = new ArrayList<>();
    private List<List<String>> ans = new ArrayList<>();

    public List<List<String>> partition(String s) {
        n = s.length();
        f = new boolean[n][n];
        for (int i = 0; i < n; ++i) {
            Arrays.fill(f[i], true);
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                f[i][j] = s.charAt(i) == s.charAt(j) && f[i + 1][j - 1];
            }
        }
        this.s = s;
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        if (i == s.length()) {
            ans.add(new ArrayList<>(t));
            return;
        }
        for (int j = i; j < n; ++j) {
            if (f[i][j]) {
                t.add(s.substring(i, j + 1));
                dfs(j + 1);
                t.remove(t.size() - 1);
            }
        }
    }
}