blocks|key|2825135|text|Java代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2825136|int[]+arr+=+{1,2,3,4,5,6,8,10,11};
int+start+=+arr[0],+last+=+arr[0];
String+output+=+"";

for+(int+i+=+1;+i+<=+arr.length;+i%2B%2B)
{
++if+(i+==+arr.length+%7C%7C+arr[i]+!=+last%2B1)
++{
++++if+(output.length()+!=+0)
++++++output+%2B=+",";
++++if+(start+==+last)
++++++output+%2B=+start;
++++else
++++++output+%2B=+start+%2B+"-"+%2B+last;
++++if+(i+!=+arr.length)
++++++start+=+last+=+arr[i];
++}
++else
+++++last+=+arr[i];
}

System.out.println(output);|code-block|syntax|javascript|2825137|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|$]]

Java code:

<pre><code>int[] arr = {1,2,3,4,5,6,8,10,11};
int start = arr[0], last = arr[0];
String output = "";

for (int i = 1; i &lt;= arr.length; i++)
{
 if (i == arr.length || arr[i] != last+1)
 {
 if (output.length() != 0)
 output += ",";
 if (start == last)
 output += start;
 else
 output += start + "-" + last;
 if (i != arr.length)
 start = last = arr[i];
 }
 else
 last = arr[i];
}

System.out.println(output);
</code></pre>

blocks|key|197154|text|这里有一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|197155|++++++++int[]+numbers+=+{+1,+2,+3,+4,+5,+6,+8,+10,+11+};

++++++++int+start,+end;
++++++++for+(int+i+=+0;+i+<+numbers.Length;+i%2B%2B)
++++++++{
++++++++++++start+=+numbers[i];

++++++++++++while+(i+<+numbers.Length+-+1+&&+numbers[i]+%2B+1+==+numbers[i+%2B+1])
++++++++++++++++i%2B%2B;

++++++++++++end+=+numbers[i];

++++++++++++if(start+==+end)
++++++++++++++++Console.WriteLine(start);
++++++++++++else
++++++++++++++++Console.WriteLine(start+%2B+"+-+"+%2B+end);
++++++++}|code-block|syntax|javascript|197156|这将显示随范围递增的后续数字。非线性递增的数字不会作为范围的一部分写入。|197157|这是第一种方法的另一个版本，它使用相同的for循环来迭代范围：|offset|length|style|CODE|197158|++++++++int+temp+=+numbers[0],+start,+end;
++++++++for+(int+i+=+0;+i+<+numbers.Length;+i%2B%2B)
++++++++{
++++++++++++start+=+temp;

++++++++++++if+(i+<+numbers.Length+-+1+)
++++++++++++++++//+if+subsequent+numbers+are+incremental+loop+further
++++++++++++++++if+(numbers[i]+%2B+1+==+numbers[i+%2B+1])
++++++++++++++++++++continue;
++++++++++++++++//+if+they+are+not,+number+at+index+i+%2B+1+is+a+new+'start'+for+the+next+iteration
++++++++++++++++else
++++++++++++++++++++temp+=+numbers[i+%2B+1];

++++++++++++end+=+numbers[i];

++++++++++++if+(start+==+end)
++++++++++++++++Console.WriteLine(start);
++++++++++++else
++++++++++++++++Console.WriteLine(start+%2B+"+-+"+%2B+end);
++++++++}|197159|entityMap^0|0|0|0|K|3|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Here is one way of doing it:

<pre><code> int[] numbers = { 1, 2, 3, 4, 5, 6, 8, 10, 11 };

 int start, end;
 for (int i = 0; i &lt; numbers.Length; i++)
 {
 start = numbers[i];

 while (i &lt; numbers.Length - 1 &amp;&amp; numbers[i] + 1 == numbers[i + 1])
 i++;

 end = numbers[i];

 if(start == end)
 Console.WriteLine(start);
 else
 Console.WriteLine(start + " - " + end);
 }
</code></pre>

This will display subsequent numbers that grow incrementally as range. Numbers that are not increasing linearly are not written as part of a range.

Here is another version of the first approach, it utilizes the same <code>for</code> loop to iterate on range:

<pre><code> int temp = numbers[0], start, end;
 for (int i = 0; i &lt; numbers.Length; i++)
 {
 start = temp;

 if (i &lt; numbers.Length - 1 )
 // if subsequent numbers are incremental loop further
 if (numbers[i] + 1 == numbers[i + 1])
 continue;
 // if they are not, number at index i + 1 is a new 'start' for the next iteration
 else
 temp = numbers[i + 1];

 end = numbers[i];

 if (start == end)
 Console.WriteLine(start);
 else
 Console.WriteLine(start + " - " + end);
 }
</code></pre>

blocks|key|199741|text|C#中的一个简单实现可能如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|199742|public+string+Format(IEnumerable<int>+input)
{
++++var+result+=+string.Empty;

++++var+previous+=+-1;
++++var+start+=+-1;
++++var+first+=+true;

++++foreach(var+i+in+input)
++++{
++++++++if(start+==+-1)
++++++++++++start+=+i;
++++++++else+if(previous+%2B+1+!=+i)
++++++++{
++++++++++++result+%2B=+FormatRange(start,+previous,+first);
++++++++++++first+=+false;
++++++++++++start+=+i;
++++++++}

++++++++previous+=+i;
++++}

++++if(start+!=+-1)
++++++++result+%2B=+FormatRange(start,+previous,+first);

++++return+result;
}

public+string+FormatRange(int+start,+int+end,+bool+isFirst)
{
++++var+result+=+string.Empty;
++++if(!isFirst)
++++++++result+%2B=+",+";
++++if(start+==+end)
++++++++result+%2B=+start;
++++else
++++++++result+%2B=+string.Format("{0}-{1}",+start,+end);
++++return+result;
}|code-block|syntax|javascript|199743|这也将为输入1,2,3输出1-3，这是完全有效的。如果没有说明，输出应该是什么，就不可能回答这一部分。|offset|length|style|CODE|199744|entityMap^0|0|0|6|5|D|3|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]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

A simple implementation in C# could look like this:

<pre><code>public string Format(IEnumerable&lt;int&gt; input)
{
 var result = string.Empty;

 var previous = -1;
 var start = -1;
 var first = true;

 foreach(var i in input)
 {
 if(start == -1)
 start = i;
 else if(previous + 1 != i)
 {
 result += FormatRange(start, previous, first);
 first = false;
 start = i;
 }

 previous = i;
 }

 if(start != -1)
 result += FormatRange(start, previous, first);

 return result;
}

public string FormatRange(int start, int end, bool isFirst)
{
 var result = string.Empty;
 if(!isFirst)
 result += ", ";
 if(start == end)
 result += start;
 else
 result += string.Format("{0}-{1}", start, end);
 return result;
}
</code></pre>

This will also output <code>1-3</code> for the input <code>1,2,3</code>, which is perfectly valid. Without a specification what the output should be instead it's impossible to answer that part.

blocks|key|3038058|text|这不是有效的C#代码，只是为了展示想法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3038059|从最小到最大对列表进行排序，然后执行以下操作：|3038060|For+i+=+Min+to+Max
{
++if+i+<+MaxFound
++++continue;

++int+step+=+1;
++Output+=+i;
++while+Found(i+%2B+Step)
++{
+++++Step%2B%2B;
+++++MaxFound+=+i+%2B+Step;
++}
++if+i+<+MaxFound+
++++Output+=+(i+%2B+"-"+%2B+MaxFound);

++Output+%2B=+",+";
}|code-block|syntax|javascript|3038061|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is no valid C# code but to show the Idea.

Sort the list from Min to Max then do this:

<pre><code>For i = Min to Max
{
 if i &lt; MaxFound
 continue;

 int step = 1;
 Output = i;
 while Found(i + Step)
 {
 Step++;
 MaxFound = i + Step;
 }
 if i &lt; MaxFound 
 Output = (i + "-" + MaxFound);

 Output += ", ";
}
</code></pre>

blocks|key|199766|text|以下是其中一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|199767|public+static+void+main(String[]+args)+{
++++print(1,+2,+3,+4,+5,+7,+9,+10,+12);
}

public+static+void+print(int+...+nums)+{
++++System.out.print(nums[0]);
++++int+idx+=+1;

++++for(int+i+=+1;+i+<+nums.length;+i%2B%2B,+idx%2B%2B)+{
++++++++if(nums[i]+-+nums[i+-+1]+!=+1)+{
++++++++++++if(idx+>+1)+{
++++++++++++++++System.out.print("+-+"+%2B+nums[i+-+1]);
++++++++++++}
++++++++++++System.out.print(",+"+%2B+nums[i]);
++++++++++++idx+=+0;
++++++++}
++++}

++++if(idx+>+1)
++++++++System.out.println("+-+"+%2B+nums[nums.length+-+1]);
}|code-block|syntax|javascript|199768|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|$]]

Here is one of the approach:

<pre><code>public static void main(String[] args) {
 print(1, 2, 3, 4, 5, 7, 9, 10, 12);
}

public static void print(int ... nums) {
 System.out.print(nums[0]);
 int idx = 1;

 for(int i = 1; i &lt; nums.length; i++, idx++) {
 if(nums[i] - nums[i - 1] != 1) {
 if(idx &gt; 1) {
 System.out.print(" - " + nums[i - 1]);
 }
 System.out.print(", " + nums[i]);
 idx = 0;
 }
 }

 if(idx &gt; 1)
 System.out.println(" - " + nums[nums.length - 1]);
}
</code></pre>

blocks|key|2825276|text|这是我最好的尝试。不是很聪明，但我相信它足够简单，可以满足这个要求。我仍然很困惑为什么"1-3“是错误的……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2825277|++++var+numbers+=+new+int[]+{+1,+2,+3,+4,+5,+6,+8,+10,+11,+12+};

++++var+groups+=+new+Dictionary<int,+int>();
++++groups.Add(numbers.First(),+numbers.First());

++++foreach+(var+num+in+numbers.Skip(1))
++++{
++++++++var+grp+=+groups.Last();
++++++++if+(grp.Value+%2B+1+==+num)
++++++++{
++++++++++++groups[grp.Key]+=+num;
++++++++}
++++++++else
++++++++{
++++++++++++groups.Add(num,+num);
++++++++}
++++}

++++var+output+=+string.Join(",",+groups.Select(grp+=>+(grp.Key+==+grp.Value)+?+grp.Value.ToString()+:+grp.Key.ToString()+%2B+"-"+%2B+grp.Value.ToString()));|code-block|syntax|javascript|2825278|注意:当然，使用字典和linq等是完全没有必要的(对于需要算法的答案来说，也太具体了)，但我认为它很好地突出了问题的分组方面|2825279|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|$]]

Heres my best attempt. Not clever, but simple enough to satisfy that requirement I believe. I'm still pretty confused as to why "1-3" was wrong though....

<pre><code> var numbers = new int[] { 1, 2, 3, 4, 5, 6, 8, 10, 11, 12 };

 var groups = new Dictionary&lt;int, int&gt;();
 groups.Add(numbers.First(), numbers.First());

 foreach (var num in numbers.Skip(1))
 {
 var grp = groups.Last();
 if (grp.Value + 1 == num)
 {
 groups[grp.Key] = num;
 }
 else
 {
 groups.Add(num, num);
 }
 }

 var output = string.Join(",", groups.Select(grp =&gt; (grp.Key == grp.Value) ? grp.Value.ToString() : grp.Key.ToString() + "-" + grp.Value.ToString()));
</code></pre>

Note: of course using the dictionary and linq etc is completely unnecessary (and way too specific for an answer requiring an algorithm), but I thought it highlighted the grouping aspect of the problem nicely

blocks|key|3038154|text|下面对连续的整数进行分组，并为每组输出一个字符串。但是，它还允许您指定要用连字符连接的组的最小长度；任何小于此长度的值都只会给出单个数字。因此，如果只想用连字符连接4或更多的组，可以传入4；如果要用连字符连接对，可以传入2。(我自己也想用3，但我不知道他们想要什么。)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3038155|它在运行过程中也不保存任何数字集合，因为您不需要这样做。|3038156|方法：|3038157|static+IEnumerable<string>+Group(IEnumerable<int>+input,+int+minLength)
{
++++int+currentStart+=+int.MinValue;
++++int+currentLength+=+0;
++++foreach+(int+c+in+input)
++++{
++++++++if+(currentLength+>+0)
++++++++++++if+(currentStart+%2B+currentLength+==+c)
++++++++++++++++currentLength%2B%2B;
++++++++++++else
++++++++++++{
++++++++++++++++if+(currentLength+>=+minLength)
++++++++++++++++++++yield+return+string.Format("{0}-{1}",
++++++++++++++++++++++++currentStart,+currentStart+%2B+currentLength+-+1);
++++++++++++++++else
++++++++++++++++++++for+(int+i+=+currentStart;+i+<+currentStart+%2B+currentLength;+i%2B%2B)
++++++++++++++++++++++++yield+return+i.ToString();
++++++++++++++++currentStart+=+c;
++++++++++++++++currentLength+=+1;
++++++++++++}
++++++++else
++++++++{
++++++++++++currentStart+=+c;
++++++++++++currentLength+=+1;
++++++++}
++++}
++++if+(currentLength+>=+minLength)
++++++++yield+return+string.Format("{0}-{1}",
++++++++++++currentStart,+currentStart+%2B+currentLength+%2B+1);
++++else
++++++++for+(int+i+=+currentStart;+i+<+currentStart+%2B+currentLength;+i%2B%2B)
++++++++++++yield+return+i.ToString();
}|code-block|syntax|javascript|3038158|用法：|3038159|int+minCount+=+3;
int[]+input+=+new[]+{+1,+2,+3,+4,+5,+6,+8,+10,+11+};
Console.WriteLine(String.Join(",",+Group(input,+minCount)));|3038160|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The following groups consecutive integers, and outputs a string for each group. However, it also allows you to specify the minimum length of group which you want to hyphenate; anything less will just give you the individual numbers. Thus if you only want to hyphenate groups of 4 or more, you can pass in 4; if you want to hyphenate pairs, you can pass in 2. (I'd want to use 3 myself, but I can't tell what they want.)

It also doesn't keep any collections of numbers as it goes along, because you don't need to.

Method:

<pre><code>static IEnumerable&lt;string&gt; Group(IEnumerable&lt;int&gt; input, int minLength)
{
 int currentStart = int.MinValue;
 int currentLength = 0;
 foreach (int c in input)
 {
 if (currentLength &gt; 0)
 if (currentStart + currentLength == c)
 currentLength++;
 else
 {
 if (currentLength &gt;= minLength)
 yield return string.Format("{0}-{1}",
 currentStart, currentStart + currentLength - 1);
 else
 for (int i = currentStart; i &lt; currentStart + currentLength; i++)
 yield return i.ToString();
 currentStart = c;
 currentLength = 1;
 }
 else
 {
 currentStart = c;
 currentLength = 1;
 }
 }
 if (currentLength &gt;= minLength)
 yield return string.Format("{0}-{1}",
 currentStart, currentStart + currentLength + 1);
 else
 for (int i = currentStart; i &lt; currentStart + currentLength; i++)
 yield return i.ToString();
}
</code></pre>

Usage:

<pre><code>int minCount = 3;
int[] input = new[] { 1, 2, 3, 4, 5, 6, 8, 10, 11 };
Console.WriteLine(String.Join(",", Group(input, minCount)));
</code></pre>

blocks|key|199804|text|可能不是面试问题的合适答案，但使用LINQ是解决这个问题的另一种方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|199805|int[]+numbers+=+{+1,+2,+3,+4,+5,+6,+8,+10,+11+};
var+remains+=+numbers.AsEnumerable();

while+(remains.Any())
{
++++int+first+=+remains.First();
++++int+last+=+remains.TakeWhile((x,+i)+=>+x+-+first+==+i).Last();
++++remains+=+remains.Skip(last+-+first+%2B+1);
++++Console.Write(first+%2B+(first+==+last+?+""+:+"-"+%2B+last)+%2B+(remains.Any()+?+","+:+Environment.NewLine));
}|code-block|syntax|javascript|199806|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|$]]

Probably not a suitable answer for an interview question, but using LINQ is another way to solve this.

<pre><code>int[] numbers = { 1, 2, 3, 4, 5, 6, 8, 10, 11 };
var remains = numbers.AsEnumerable();

while (remains.Any())
{
 int first = remains.First();
 int last = remains.TakeWhile((x, i) =&gt; x - first == i).Last();
 remains = remains.Skip(last - first + 1);
 Console.Write(first + (first == last ? "" : "-" + last) + (remains.Any() ? "," : Environment.NewLine));
}
</code></pre>

blocks|key|197993|text|这是一个Haskell版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|197994|import+Data.List

parseRange+[]+=+""
parseRange+n+=+
++let+range+=+takeWhile+(\x+->+isInfixOf+[x,x%2B1]+n)+n
++in+if+not+(null+range)
++++++++then+show+(head+range)+%2B%2B+"-"+%2B%2B+show+(last+range+%2B+1)+
+++++++++++++%2B%2B+(if+length+(tail+n)+>+1+then+","+else+"")+
+++++++++++++%2B%2B+parseRange+(drop+(length+range+%2B+1)+n)+
++++++++else+show+(head+n)+%2B%2B+(if+null+(tail+n)+then+""+else+",")+
+++++++++++++%2B%2B+parseRange+(drop+1+n)|code-block|syntax|javascript|197995|输出：|197996|*Main>+parseRange+[1,2,3,4,5,6,8,10,11]
"1-6,8,10-11"|197997|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here's a Haskell version:

<pre><code>import Data.List

parseRange [] = ""
parseRange n = 
 let range = takeWhile (\x -&gt; isInfixOf [x,x+1] n) n
 in if not (null range)
 then show (head range) ++ "-" ++ show (last range + 1) 
 ++ (if length (tail n) &gt; 1 then "," else "") 
 ++ parseRange (drop (length range + 1) n) 
 else show (head n) ++ (if null (tail n) then "" else ",") 
 ++ parseRange (drop 1 n)
</code></pre>

Output:

<pre><code>*Main&gt; parseRange [1,2,3,4,5,6,8,10,11]
"1-6,8,10-11"
</code></pre>

blocks|key|199822|text|还有一种在F#中使用fold的方法--只是为了好玩。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|199823|let+parseRange+numbers+=+
++numbers+
++%7C>+Seq.fold+
++++(fun+list+n+->+
++++++match+list+with
++++++%7C(a,b)+::+tail+when+b%2B1+=+n+->+(a,+n)+::+tail
++++++%7C_+->+(n,n)+::+list)+[]
++%7C>+List.rev+
++%7C>+Seq.map+(fun+(a,b)+->+if+a+=+b+then+sprintf+"%25i"+a+else+sprintf+"%25i-%25i"+a+b)
++%7C>+String.concat+","|code-block|syntax|javascript|199824|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|$]]

And a way to do it with fold in F# - just for fun.

<pre><code>let parseRange numbers = 
 numbers 
 |&gt; Seq.fold 
 (fun list n -&gt; 
 match list with
 |(a,b) :: tail when b+1 = n -&gt; (a, n) :: tail
 |_ -&gt; (n,n) :: list) []
 |&gt; List.rev 
 |&gt; Seq.map (fun (a,b) -&gt; if a = b then sprintf "%i" a else sprintf "%i-%i" a b)
 |&gt; String.concat ","
</code></pre>

<blockquote>
 I have this sequence 1,2,3,4,5,6,8,10,11
 
 Expected output is 1-6,8,10-11
</blockquote>

This problem is about formatting the sequence in easy readable form

I tried with c# and used many if &amp; else. 

Interviewer said, there is some simple algorithm to do this.

I have no idea how to achive this very simple.

<blockquote>
 Also for 1,2,3 i shown 1-3. They said its wrong!.
</blockquote>

Is there any design pattern(interpreter) involved in this logic?

Display 1,2,3,4,5,6,8,10,11 as 1-6,8,10-11

解释器

Java

我有这个序列1,2,3,4,5,6,8,10,11，我有这个序列预期输出为1-6,8,10-11这个问题是关于以易于阅读的形式格式化序列我尝试了c#，并使用了很多if & else。面试官说，有一些简单的算法可以做到这一点。我不知道如何简单地实现这一点。对于1,2,3，我也显示了1-3的。他们说错了！这个逻辑中是否涉及...

问将1,2,3,4,5,6,8,10,11显示为1-6,8,10-11
EN

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

问将1,2,3,4,5,6,8,10,11显示为1-6,8,10-11EN

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

问将1,2,3,4,5,6,8,10,11显示为1-6,8,10-11
EN