<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 {
public:
static bool sortbydig(int &a,int &b){
    string x=to_string(a);
    string y=to_string(b);
    if((int)x.size()<=(int)y.size()){
        bool p=true;
        for(int i=0;i<(int)x.size();i++){
            if(x[i]!=y[i]){
                if((int)x[i]>(int)y[i]){
                    p=false;
                }
                break;
            }
        }
        return p;
    }
    else{
            bool p=false;
        for(int i=0;i<(int)y.size();i++){
            if(x[i]!=y[i]){
                if((int)x[i]<(int)y[i]){
                    p=true;
                }
                break;
            }
        }
        return p;
    }
}
    vector<int> lexicalOrder(int n){
        vector<int>v;
        for(int i=0;i<n;i++){
            v.push_back(i+1);
        }
        sort(v.begin(),v.end(),sortbydig);
        return v;
    }
};
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