blocks|key|189695|text|def+has22(nums):
++++return+any(x+==+y+==+2+for+x,+y+in+zip(nums,+nums[1:]))

>>>+has22([1,+2,+2])
True
>>>+has22([1,+2,+1,+2])
False
>>>+has22([2,+1,+2])
False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|189696|在Python2中，如果需要惰性zip，请使用：from+itertools+import+izip|unstyled|offset|length|style|CODE|189697|entityMap^0|0|G|3|O|Q|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>def has22(nums):
 return any(x == y == 2 for x, y in zip(nums, nums[1:]))

&gt;&gt;&gt; has22([1, 2, 2])
True
&gt;&gt;&gt; has22([1, 2, 1, 2])
False
&gt;&gt;&gt; has22([2, 1, 2])
False
</code></pre>

In Python 2 use: <code>from itertools import izip</code> if you want a lazy <code>zip</code>

blocks|key|3018039|text|可能是最简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3018040|def+has22(nums):
++++return+(2,+2)+in+zip(nums,+nums[1:])|code-block|syntax|javascript|3018041|3018042|假设nums+==+[1,+2,+2,+3,+4,+5]。然后对nums[1:]+==+[2,+2,+3,+4,+5]执行follows操作。当作为zip(nums,+nums[1:])调用时，zip()函数将它们压缩到具有相同元组的迭代器中，如下所示：|offset|length|style|CODE|3018043|nums++++++=>++[1,++++++2,++++++2,++++++3,++++++4,++++++5]
nums[1:]++=>++[2,++++++2,++++++3,++++++4,++++++5]
zip()+++++=>++[(1,+2),+(2,+2),+(2,+3),+(3,+4),+(4,+5)]|3018044|而且应该很清楚(2,+2)+in+[(1,+2),+(2,+2),+(2,+3),+(3,+4),+(4,+5)]是如何实现的。这与(2,+2)+in+zip(nums,+nums[1:])的结果相同。|3018045|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/functions.html#slice|1|https://docs.python.org/3/library/functions.html#zip^0|0|0|0|2|Q|W|R|22|J|1P|7|0|2P|5|1|0|0|7|1E|1U|T|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|14|8|@$J|15|K|16|L|M]|$J|17|K|18|L|M]|$J|19|K|1A|L|M]]|9|@$J|1B|K|1C|1|1D]|$J|1E|K|1F|1|1G]]|A|$]]|$1|N|3|O|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|P|3|Q|5|6|7|1I|8|@$J|1J|K|1K|L|M]|$J|1L|K|1M|L|M]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y]]|Z|$5|U|V|W|A|$X|10]]]]

Potentially the simplest solution:

<pre><code>def has22(nums):
 return (2, 2) in zip(nums, nums[1:])
</code></pre>

<hr>

Suppose <code>nums == [1, 2, 2, 3, 4, 5]</code>. It then <a href="https://docs.python.org/3/library/functions.html#slice" rel="noreferrer">follows</a> that <code>nums[1:] == [2, 2, 3, 4, 5]</code>. The <a href="https://docs.python.org/3/library/functions.html#zip" rel="noreferrer">zip()</a> function, when called as <code>zip(nums, nums[1:])</code>, zips them into an iterator of the same tuples as below:

<pre><code>nums =&gt; [1, 2, 2, 3, 4, 5]
nums[1:] =&gt; [2, 2, 3, 4, 5]
zip() =&gt; [(1, 2), (2, 2), (2, 3), (3, 4), (4, 5)]
</code></pre>

And it should be clear how <code>(2, 2) in [(1, 2), (2, 2), (2, 3), (3, 4), (4, 5)]</code> is true. This is the same result as <code>(2, 2) in zip(nums, nums[1:])</code>.

blocks|key|187495|text|def+has22(nums):
++++for+x+in+range(len(nums)-1):
++++++++if+(nums[x]+==+2)+and+(nums[x%2B1]+==+2):
++++++++++++return+True
++++return+False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|187496|我刚刚更正了你的代码。它的运行时间是线性的，所以没有任何理由去进一步研究它。|unstyled|187497|下面是codebunk上的运行代码。http://codebunk.com/bunk#-Ivk7Xw2blX3cIWavI17|offset|length|187498|entityMap|0|LINK|mutability|MUTABLE|url|http://codebunk.com/bunk#-Ivk7Xw2blX3cIWavI17^0|0|0|I|19|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<pre><code>def has22(nums):
 for x in range(len(nums)-1):
 if (nums[x] == 2) and (nums[x+1] == 2):
 return True
 return False
</code></pre>

I have just corrected your code. It runs in linear time so don't see any reason to work on it further.

Here is the running code on codebunk. <a href="http://codebunk.com/bunk#-Ivk7Xw2blX3cIWavI17" rel="noreferrer">http://codebunk.com/bunk#-Ivk7Xw2blX3cIWavI17</a>

blocks|key|3017921|text|您可以使用iter()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3017922|>>>+def+has22(lst):
...+++++lst+=+iter(lst)
...+++++for+i+in+lst:
...+++++++++++++try:
...+++++++++++++++++if+i+==+2+and+lst.next()+==+2:
...+++++++++++++++++++++return+True
...+++++++++++++except+StopIteration:
...+++++++++++++++++++++pass
...+++++return+False
...+
>>>+has22([1,+2,+2])
True
>>>+has22([1,+2,+1,+2])
False
>>>+has22([2,+1,+2])
False|code-block|syntax|javascript|3017923|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/functions.html#iter^0|5|6|5|6|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

You can use <a href="http://docs.python.org/2/library/functions.html#iter" rel="nofollow"><code>iter()</code></a>:

<pre><code>&gt;&gt;&gt; def has22(lst):
... lst = iter(lst)
... for i in lst:
... try:
... if i == 2 and lst.next() == 2:
... return True
... except StopIteration:
... pass
... return False
... 
&gt;&gt;&gt; has22([1, 2, 2])
True
&gt;&gt;&gt; has22([1, 2, 1, 2])
False
&gt;&gt;&gt; has22([2, 1, 2])
False
</code></pre>

blocks|key|3017951|text|def+has22(nums):
++++it+=+iter(nums)
++++return+any(x+==+2+==+next(it,+None)+for+x+in+it)

>>>+has22([1,+2,+2])
True
>>>+has22([1,+2,+1,+2])
False
>>>+has22([2,+1,+2])
False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3017952|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def has22(nums):
 it = iter(nums)
 return any(x == 2 == next(it, None) for x in it)

&gt;&gt;&gt; has22([1, 2, 2])
True
&gt;&gt;&gt; has22([1, 2, 1, 2])
False
&gt;&gt;&gt; has22([2, 1, 2])
False
</code></pre>

blocks|key|187549|text|def+has22(nums):
++++for+i+in+range(len(nums)+-+1):
++++++++if+nums[i]+==+2+and+nums[i+%2B+1]+==+2:
++++++++++++return+True
++++return+False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|187550|这是我想出的最简单的解决方案。|unstyled|187551|使用for循环检查迭代数，“nums[i]+==+2”和它旁边的那个，也是[i%2B1]+==+2。|offset|length|style|CODE|187552|(len(nums)-1)：这一行阻止它通过for循环超出范围，因为最后一个循环上的i%2B1将检查出范围。|187553|entityMap^0|0|0|E|C|11|A|0|0|D|16|3|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|F|7|X|8|@$I|Y|J|Z|K|L]|$I|10|J|11|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|F|7|12|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>def has22(nums):
 for i in range(len(nums) - 1):
 if nums[i] == 2 and nums[i + 1] == 2:
 return True
 return False
</code></pre>

This was the simplest solution I came up with. 

Using a for loop to check if the iterated number, <code>nums[i] == 2</code> "and" the one very next to it, <code>[i+1] == 2</code> as well.

<code>(len(nums)-1)</code>: this line prevents it from going out of the range through the for loop as the <code>i+1</code> on the final loop will check out of the range.

blocks|key|2805818|text|def+has22(nums):
++++return+'22'+in+''.join(map(str,+nums))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2805819|列表1、2、2+->字符串“122”|unstyled|2805820|在'122'中返回'22'|offset|length|style|CODE|2805821|entityMap^0|0|0|1|5|9|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>def has22(nums):
 return '22' in ''.join(map(str, nums))
</code></pre>

list [1, 2, 2] -> str '122' 
return <code>'22'</code> in <code>'122'</code>

blocks|key|3017915|text|使用enumerate()来获取索引和项目，遍历list只返回元素，而不返回索引。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3017916|def+has22(lis):
++++for+i,x+in+enumerate(lis[:-1]):
+++++++if+x==2+and+lis[i%2B1]==2:
+++++++++++return+True
++++return+False

>>>+has22([1,+2,+2])+
True
>>>+has22([1,+2,+1,+2])
False
>>>+has22([2,+1,+2])
False|code-block|syntax|javascript|3017917|entityMap^0|2|B|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Use <code>enumerate()</code> to get both index as well as item, iterating over list only returns it's elements not index.

<pre><code>def has22(lis):
 for i,x in enumerate(lis[:-1]):
 if x==2 and lis[i+1]==2:
 return True
 return False

&gt;&gt;&gt; has22([1, 2, 2]) 
True
&gt;&gt;&gt; has22([1, 2, 1, 2])
False
&gt;&gt;&gt; has22([2, 1, 2])
False
</code></pre>

blocks|key|186977|text|我会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|186978|def+has22(l):
++++return+any(l[i]==2+and+l[i%2B1]==2+for+i+in+xrange(len(l)-1))|code-block|syntax|javascript|186979|这使用了与其他答案类似的想法，但与生成器一起工作(在这样的情况下会更好)。|186980|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I would do this:

<pre><code>def has22(l):
 return any(l[i]==2 and l[i+1]==2 for i in xrange(len(l)-1))
</code></pre>

This uses a similar idea as the other answers, but works with a generator (as would be preferred in cases like this).

blocks|key|3017975|text|def+has22(lst):
++pos+=+0+
++while+True:
++++try:
++++++next+=+lst.index(2,+pos)+%2B+1
++++except+ValueError:
++++++return+False
++++if+next+==+pos+%2B+1:
++++++return+True
++++pos+=+next++++|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3017976|这使用了index()可能更快的想法，因为它不是用Python语言实现的。不过，并没有测量它。|unstyled|offset|length|style|CODE|3017977|关于您的问题:您的代码由于在for循环初始化时没有使用range()而受到影响。按照您的说法，x将不是索引，而是列表的元素。此外，它还需要使用=进行比较(实际上就是赋值)。使用==进行比较。|3017978|这不是一个图问题，这是一个简单的搜索问题。对于finding+strings+in+strings+(您实际要做的事情)，有非常好的strstr解决方案(除了直接的解决方案)。|3017979|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Boyer%25E2%2580%2593Moore%25E2%2580%2593Horspool_algorithm^0|0|4|7|0|E|3|R|7|1B|1|1Z|1|2G|2|0|1U|6|N|Q|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@$G|Y|H|Z|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@$G|11|H|12|I|J]|$G|13|H|14|I|J]|$G|15|H|16|I|J]|$G|17|H|18|I|J]|$G|19|H|1A|I|J]]|9|@]|A|$]]|$1|M|3|N|5|F|7|1B|8|@$G|1C|H|1D|I|J]]|9|@$G|1E|H|1F|1|1G]]|A|$]]|$1|O|3|-4|5|F|7|1H|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

<pre><code>def has22(lst):
 pos = 0 
 while True:
 try:
 next = lst.index(2, pos) + 1
 except ValueError:
 return False
 if next == pos + 1:
 return True
 pos = next 
</code></pre>

This uses the idea that <code>index()</code> might be faster due to being implemented not in Python. Didn't measure it, though.

Concerning your questions: Your code suffers from not using <code>range()</code> at the <code>for</code> loop init. The way you put it, <code>x</code> will not be the indexes but the elements of your list. And it also suffers from using <code>=</code> for comparison (which actually just is assignment). Use <code>==</code> for comparison.

This is not a graph problem, it is a simple search issue. There are quite nifty <code>strstr</code> solutions (besides the straight-forward one) for <a href="http://en.wikipedia.org/wiki/Boyer%E2%80%93Moore%E2%80%93Horspool_algorithm" rel="nofollow">finding strings in strings</a> (what you actually do).

blocks|key|185503|text|def+has22(nums):+++for+i+in+range(len(nums)-1):+++++if+nums[i:i%2B2]+==+[2,2]:+++++++return+True+++return+False|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|185504|entityMap^0|0|31|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>def has22(nums):
 for i in range(len(nums)-1):
 if nums[i:i+2] == [2,2]:
 return True
 return False</code>

blocks|key|2805740|text|++++def+has22(nums):
++++++if+len(nums)==0:
++++++++return+False
++++++for+i+in+range(len(nums)-1):
++++++++#print+nums[i:i%2B2]
++++++++if+nums[i:i%2B2]==[2,2]:
++++++++++return+True
++++++return+False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2805741|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> def has22(nums):
 if len(nums)==0:
 return False
 for i in range(len(nums)-1):
 #print nums[i:i+2]
 if nums[i:i+2]==[2,2]:
 return True
 return False
</code></pre>

blocks|key|185533|text|def+has_22(nums):
++++desired+=+[2,+2]
++++if+str(desired)[1:-1]+in+str(nums):
++++++++return+True
++++else:
++++++++return+False
++++pass|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|185534|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def has_22(nums):
 desired = [2, 2]
 if str(desired)[1:-1] in str(nums):
 return True
 else:
 return False
 pass
</code></pre>

blocks|key|189733|text|仅使用索引|type|unstyled|depth|inlineStyleRanges|entityRanges|data|189734|def+has22(nums):
++++return+nums[nums.index(2)%2B1]+==+2|code-block|syntax|javascript|189735|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Just using index

<pre><code>def has22(nums):
 return nums[nums.index(2)+1] == 2
</code></pre>

blocks|key|185536|text|def+two_two(nums):
++if+nums.count(2)==2:
++++return+True
++else:
++++return+False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|185537|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-py prettyprint-override"><code>def two_two(nums):
 if nums.count(2)==2:
 return True
 else:
 return False
</code></pre>

blocks|key|2805823|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2805824|>>>+ls+=+[1,+2,+2]
>>>+s+=+str(ls)
>>>+'2,+2'+in+s
>>>+True|code-block|syntax|javascript|2805825|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this

<pre><code>&gt;&gt;&gt; ls = [1, 2, 2]
&gt;&gt;&gt; s = str(ls)
&gt;&gt;&gt; '2, 2' in s
&gt;&gt;&gt; True

</code></pre>

blocks|key|187101|text|下面的方法对我很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|187102|def+has22(nums):
+++++for+i+in+range(0,len(nums)-1):
++++++++if+nums[i:i%2B2]+==+[2,2]
+++++++++++return+True
+++++return+False|code-block|syntax|javascript|187103|此函数检查输入的每2个值块，以查看它是否包含[2,2]|offset|length|style|CODE|187104|entityMap^0|0|0|M|5|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The following works for me:
<pre class="lang-py prettyprint-override"><code>def has22(nums):
 for i in range(0,len(nums)-1):
 if nums[i:i+2] == [2,2]
 return True
 return False
</code></pre>
this function checks every 2 value chunk of the input to see if it contains <code>[2,2]</code>

blocks|key|3018125|text|def+has22(nums):
++
++
++for+i+in+range(len(nums)-1):
++++if+nums[i]+==+2+and+nums[i%2B1]+==+2:
++++++return+True
++return+False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3018126|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def has22(nums):
 
 
 for i in range(len(nums)-1):
 if nums[i] == 2 and nums[i+1] == 2:
 return True
 return False
</code></pre>

blocks|key|189718|text|def+has22(nums):
++++numbers+=+str(nums)
++++++++if+numbers.count('2,+2')+>=1:
++++++++++++return+True
++++++++else:
++++++++++++return+False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|189719|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def has22(nums):
 numbers = str(nums)
 if numbers.count('2, 2') &gt;=1:
 return True
 else:
 return False
</code></pre>

blocks|key|185539|text|def+has22(nums):
++++for+x+in+range(0,+len(nums)-1):
++++++++if+nums[x]+==+2+and+nums[x%2B1]+==+2:
++++++++++++return+True

++++return+False|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|185540|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-py prettyprint-override"><code>def has22(nums):
 for x in range(0, len(nums)-1):
 if nums[x] == 2 and nums[x+1] == 2:
 return True

 return False
</code></pre>

I am stuck on this problem

<blockquote>
 Given an array of ints, return True if the array contains a 2 next to a 2 somewhere.
</blockquote>

<pre><code>has22([1, 2, 2]) → True
has22([1, 2, 1, 2]) → False
has22([2, 1, 2]) → False
</code></pre>

I know the basic idea (there are syntax errors) but I can't implement it. I would also like to know what type of problem this is, eg. graph, search?

<pre><code>def has22(nums):
for x in nums:
 if ( (nums[x] = 2) and (nums[x+1] = 2) )
 return True

return False 
</code></pre>

Find if the array contain a 2 next to a 2

我被这个问题卡住了。给定一个整数数组，如果该数组在某处的2旁边包含2，则返回True。has22([1, 2, 2]) → Truehas22([1, 2, 1, 2]) → Falsehas22([2, 1, 2]) → False我知道基本的想法(有语法错误)，但我不能实现它。我也想知道这是什么类型的问题，例如。图，搜索？def has22(nums):for x in nums:    if

问查看数组中2的旁边是否包含2
EN

回答 20

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问查看数组中2的旁边是否包含2EN

回答 20

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问查看数组中2的旁边是否包含2
EN