<p>Given two strings <code>s</code> and <code>t</code>, return <code>true</code> <em>if they are equal when both are typed into empty text editors</em>. <code>&#39;#&#39;</code> means a backspace character.</p>

<p>Note that after backspacing an empty text, the text will continue empty.</p>

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

<pre>
<strong>Input:</strong> s = &quot;ab#c&quot;, t = &quot;ad#c&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Both s and t become &quot;ac&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;ab##&quot;, t = &quot;c#d#&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> Both s and t become &quot;&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;a#c&quot;, t = &quot;b&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> s becomes &quot;c&quot; while t becomes &quot;b&quot;.
</pre>

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

<ul>
	<li><code><span>1 &lt;= s.length, t.length &lt;= 200</span></code></li>
	<li><span><code>s</code> and <code>t</code> only contain lowercase letters and <code>&#39;#&#39;</code> characters.</span></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Can you solve it in <code>O(n)</code> time and <code>O(1)</code> space?</p>

class Solution {
public:
    bool backspaceCompare(string s, string t) {
        stack<char> st;
        for (int i = 0; i < s.length(); i++) {
            if (!st.empty() && s[i] == '#')
                st.pop();
            else if (s[i] != '#')
                st.push(s[i]);
        }
        string s1 = "";
        while (!st.empty()) {
            s1 += st.top();
            st.pop();
        }
        for (int i = 0; i < t.length(); i++) {
            if (!st.empty() && t[i] == '#')
                st.pop();
            else if (t[i] != '#')
                st.push(t[i]);
        }
        string s2 = "";
        while (!st.empty()) {
            s2 += st.top();
            st.pop();
        }
        return s1 == s2;
    }
};
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