<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:
    def dayOfYear(self, date: str) -> int:
        y=int(date[0:4])
        d=0
        def leap(y,d):
            if y%400==0 or (y%4==0 and y%100!=0):
                return 1
            else:
                return 0
        if date[5:7]=="01":
            return int(date[8:10])
        if date[5:7]=="02":
            d+=31
        if date[5:7]=="03":
            d+=59
            o=leap(y,d)
            d+=o
        if date[5:7]=="04":
            d+=90
            o=leap(y,d)
            d+=o
        if date[5:7]=="05":
            d+=120
            o=leap(y,d)
            d+=o
        if date[5:7]=="06":
            d+=151
            o=leap(y,d)
            d+=o
        if date[5:7]=="07":
            d+=181
            o=leap(y,d)
            d+=o
        if date[5:7]=="08":
            d+=212
            o=leap(y,d)
            d+=o
        if date[5:7]=="09":
            d+=243
            o=leap(y,d)
            d+=o
        if date[5:7]=="10":
            d+=273
            o=leap(y,d)
            d+=o
        if date[5:7]=="11":
            d+=304
            o=leap(y,d)
            d+=o
        if date[5:7]=="12":
            d+=334
            o=leap(y,d)
            d+=o
        d+=int(date[8:10])
        return d
16600