Longest Substring Without Repeating Characters
<p>Given a string <code>s</code>, find the length of the <strong>longest</strong> <span data-keyword="substring-nonempty"><strong>substring</strong></span> without duplicate characters.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abcabcbb&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The answer is &quot;abc&quot;, with the length of 3.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;bbbbb&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> The answer is &quot;b&quot;, with the length of 1.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;pwwkew&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The answer is &quot;wke&quot;, with the length of 3.
Notice that the answer must be a substring, &quot;pwke&quot; is a subsequence and not a substring.
</pre>

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

<ul>
	<li><code>0 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>s</code> consists of English letters, digits, symbols and spaces.</li>
</ul>
Hash Table; String; Sliding Window
impl Solution {
    pub fn length_of_longest_substring(s: String) -> i32 {
        let mut cnt = [0; 128];
        let mut ans = 0;
        let mut l = 0;
        let chars: Vec<char> = s.chars().collect();
        let n = chars.len();
        for (r, &c) in chars.iter().enumerate() {
            cnt[c as usize] += 1;
            while cnt[c as usize] > 1 {
                cnt[chars[l] as usize] -= 1;
                l += 1;
            }
            ans = ans.max((r - l + 1) as i32);
        }
        ans
    }
}
