<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 repeating 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>

#include<unordered_map>

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        unordered_map<char, int> alphabet;
        alphabet.reserve(256);

        int length = 0;
        string::iterator i, j;
        for (i = s.begin(); i != s.end();) {
            alphabet.emplace(*i, 0);
            for (j = i + 1; j != s.end();) {
                if (alphabet.find(*j) == alphabet.end()) {
                    alphabet.emplace(*j, j - i);
                    ++j;
                    continue;
                }
                break;
            }
        
            if (int tempLength = j - i; length < tempLength) {
                length = tempLength;
            }

            i += alphabet[*j] + 1;
            alphabet.clear();
        }

        return length;
    }
};
150040