<p>Given an integer <code>n</code>, return <em>a string array </em><code>answer</code><em> (<strong>1-indexed</strong>) where</em>:</p>

<ul>
	<li><code>answer[i] == &quot;FizzBuzz&quot;</code> if <code>i</code> is divisible by <code>3</code> and <code>5</code>.</li>
	<li><code>answer[i] == &quot;Fizz&quot;</code> if <code>i</code> is divisible by <code>3</code>.</li>
	<li><code>answer[i] == &quot;Buzz&quot;</code> if <code>i</code> is divisible by <code>5</code>.</li>
	<li><code>answer[i] == i</code> (as a string) if none of the above conditions are true.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 3
<strong>Output:</strong> ["1","2","Fizz"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 5
<strong>Output:</strong> ["1","2","Fizz","4","Buzz"]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> n = 15
<strong>Output:</strong> ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

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

class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string>s;
        
        for(int i=1; i<=n; i++){
            if(i%3==0 && i%5==0){
                s.push_back("FizzBuzz");
            }
            else if(i%3==0){
                s.push_back("Fizz");
            }
            else if(i%5==0){
                s.push_back("Buzz");
            }
            else{
                int temp = i;
                string t = "";
                while(temp!=0){
                    int digit = temp%10;
                    t += digit+'0';
                    temp = temp/10;
                }
                reverse(t.begin(),t.end());
                s.push_back(t);
            }
        }
        return s;
    }
};
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