Reverse Words in a String
<p>Given an input string <code>s</code>, reverse the order of the <strong>words</strong>.</p>

<p>A <strong>word</strong> is defined as a sequence of non-space characters. The <strong>words</strong> in <code>s</code> will be separated by at least one space.</p>

<p>Return <em>a string of the words in reverse order concatenated by a single space.</em></p>

<p><b>Note</b> that <code>s</code> may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.</p>

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

<pre>
<strong>Input:</strong> s = &quot;the sky is blue&quot;
<strong>Output:</strong> &quot;blue is sky the&quot;
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;  hello world  &quot;
<strong>Output:</strong> &quot;world hello&quot;
<strong>Explanation:</strong> Your reversed string should not contain leading or trailing spaces.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;a good   example&quot;
<strong>Output:</strong> &quot;example good a&quot;
<strong>Explanation:</strong> You need to reduce multiple spaces between two words to a single space in the reversed string.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> contains English letters (upper-case and lower-case), digits, and spaces <code>&#39; &#39;</code>.</li>
	<li>There is <strong>at least one</strong> word in <code>s</code>.</li>
</ul>

<p>&nbsp;</p>
<p><b data-stringify-type="bold">Follow-up:&nbsp;</b>If the string data type is mutable in your language, can&nbsp;you solve it&nbsp;<b data-stringify-type="bold">in-place</b>&nbsp;with&nbsp;<code data-stringify-type="code">O(1)</code>&nbsp;extra space?</p>
Two Pointers; String
impl Solution {
    pub fn reverse_words(s: String) -> String {
        let mut words = Vec::new();
        let s: Vec<char> = s.chars().collect();
        let mut i = 0;
        let n = s.len();

        while i < n {
            while i < n && s[i] == ' ' {
                i += 1;
            }
            if i < n {
                let mut j = i;
                while j < n && s[j] != ' ' {
                    j += 1;
                }
                words.push(s[i..j].iter().collect::<String>());
                i = j;
            }
        }

        words.reverse();
        words.join(" ")
    }
}
