<p>Given a string <code>date</code> representing a <a href="https://en.wikipedia.org/wiki/Gregorian_calendar" target="_blank">Gregorian calendar</a> date formatted as <code>YYYY-MM-DD</code>, return <em>the day number of the year</em>.</p>

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

<pre>
<strong>Input:</strong> date = &quot;2019-01-09&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong> Given date is the 9th day of the year in 2019.
</pre>

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

<pre>
<strong>Input:</strong> date = &quot;2019-02-10&quot;
<strong>Output:</strong> 41
</pre>

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

<ul>
	<li><code>date.length == 10</code></li>
	<li><code>date[4] == date[7] == &#39;-&#39;</code>, and all other <code>date[i]</code>&#39;s are digits</li>
	<li><code>date</code> represents a calendar date between Jan 1<sup>st</sup>, 1900 and Dec 31<sup>th</sup>, 2019.</li>
</ul>

class Solution {
public:
    int dayOfYear(string s) {
        int months[12] = {0, 31, 59 ,90, 120, 151, 181, 212, 243, 273, 304, 334};

        int y =0, m = 0, d =0;
        for(int i =0;i<4;i++){
            y = (10*y)+(s[i] - '0');
        } 
        for(int i =5;i<7;i++){
            m= (10*m)+(s[i] - '0');
        } 
        for(int i =8;i<10;i++){
            d= (10*d)+(s[i] - '0');
        } 

        d += months[m-1];
        if((y == 2000 || (y%4==0 )) && (m>2) && y!=1900)
            ++d;
        
        return d;

    }
};
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