Restore IP Addresses
<p>A <strong>valid IP address</strong> consists of exactly four integers separated by single dots. Each integer is between <code>0</code> and <code>255</code> (<strong>inclusive</strong>) and cannot have leading zeros.</p>

<ul>
	<li>For example, <code>&quot;0.1.2.201&quot;</code> and <code>&quot;192.168.1.1&quot;</code> are <strong>valid</strong> IP addresses, but <code>&quot;0.011.255.245&quot;</code>, <code>&quot;192.168.1.312&quot;</code> and <code>&quot;192.168@1.1&quot;</code> are <strong>invalid</strong> IP addresses.</li>
</ul>

<p>Given a string <code>s</code> containing only digits, return <em>all possible valid IP addresses that can be formed by inserting dots into </em><code>s</code>. You are <strong>not</strong> allowed to reorder or remove any digits in <code>s</code>. You may return the valid IP addresses in <strong>any</strong> order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;25525511135&quot;
<strong>Output:</strong> [&quot;255.255.11.135&quot;,&quot;255.255.111.35&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0000&quot;
<strong>Output:</strong> [&quot;0.0.0.0&quot;]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;101023&quot;
<strong>Output:</strong> [&quot;1.0.10.23&quot;,&quot;1.0.102.3&quot;,&quot;10.1.0.23&quot;,&quot;10.10.2.3&quot;,&quot;101.0.2.3&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 20</code></li>
	<li><code>s</code> consists of digits only.</li>
</ul>
String; Backtracking
func restoreIpAddresses(s string) (ans []string) {
	n := len(s)
	t := []string{}
	var dfs func(int)
	dfs = func(i int) {
		if i >= n && len(t) == 4 {
			ans = append(ans, strings.Join(t, "."))
			return
		}
		if i >= n || len(t) == 4 {
			return
		}
		x := 0
		for j := i; j < i+3 && j < n; j++ {
			x = x*10 + int(s[j]-'0')
			if x > 255 || (j > i && s[i] == '0') {
				break
			}
			t = append(t, s[i:j+1])
			dfs(j + 1)
			t = t[:len(t)-1]
		}
	}
	dfs(0)
	return
}