<p>Given an integer <code>n</code>, return all the numbers in the range <code>[1, n]</code> sorted in lexicographical order.</p>

<p>You must write an algorithm that runs in&nbsp;<code>O(n)</code>&nbsp;time and uses <code>O(1)</code> extra space.&nbsp;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 13
<strong>Output:</strong> [1,10,11,12,13,2,3,4,5,6,7,8,9]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 2
<strong>Output:</strong> [1,2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class Solution:
    def dfs(self, number, result, limit, visited):
        if number <= limit and number not in visited:
            result.append(number)
            visited.add(number)
            for i in range(10):
                if int(number*10+i) <= limit:
                    self.dfs(number*10+i, result, limit, visited)
            
    def lexicalOrder(self, n):
        result = []
        visited = set()
        for i in range(1,n+1):
            if len(str(i)) < len(str(n)):
                self.dfs(i, result, n, visited)
            elif i not in visited:
                result.append(i)
        return result
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