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
func partition(s string) (ans [][]string) {
	n := len(s)
	f := make([][]bool, n)
	for i := range f {
		f[i] = make([]bool, n)
		for j := range f[i] {
			f[i][j] = true
		}
	}
	for i := n - 1; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			f[i][j] = s[i] == s[j] && f[i+1][j-1]
		}
	}
	t := []string{}
	var dfs func(int)
	dfs = func(i int) {
		if i == n {
			ans = append(ans, append([]string(nil), t...))
			return
		}
		for j := i; j < n; j++ {
			if f[i][j] {
				t = append(t, s[i:j+1])
				dfs(j + 1)
				t = t[:len(t)-1]
			}
		}
	}
	dfs(0)
	return
}