Jewels and Stones
<p>You&#39;re given strings <code>jewels</code> representing the types of stones that are jewels, and <code>stones</code> representing the stones you have. Each character in <code>stones</code> is a type of stone you have. You want to know how many of the stones you have are also jewels.</p>

<p>Letters are case sensitive, so <code>&quot;a&quot;</code> is considered a different type of stone from <code>&quot;A&quot;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> jewels = "aA", stones = "aAAbbbb"
<strong>Output:</strong> 3
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> jewels = "z", stones = "ZZ"
<strong>Output:</strong> 0
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;=&nbsp;jewels.length, stones.length &lt;= 50</code></li>
	<li><code>jewels</code> and <code>stones</code> consist of only English letters.</li>
	<li>All the characters of&nbsp;<code>jewels</code> are <strong>unique</strong>.</li>
</ul>
Hash Table; String
use std::collections::HashSet;
impl Solution {
    pub fn num_jewels_in_stones(jewels: String, stones: String) -> i32 {
        let mut s = jewels.as_bytes().iter().collect::<HashSet<&u8>>();
        let mut ans = 0;
        for c in stones.as_bytes() {
            if s.contains(c) {
                ans += 1;
            }
        }
        ans
    }
}
