blocks|key|794489|text|int[]+arr1+=+{1,3,4,9,8};
int[]+arr2+=+new+int[5];

int+j+=+0;

for(int+i+=+arr1.Length+-+1;+i+>=+0;+i--)
{
++arr2[j]+=+arr1[i];
++j%2B%2B;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|794490|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>int[] arr1 = {1,3,4,9,8};
int[] arr2 = new int[5];

int j = 0;

for(int i = arr1.Length - 1; i &gt;= 0; i--)
{
 arr2[j] = arr1[i];
 j++;
}
</code></pre>

blocks|key|794502|text|尝试如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794503|var+counter+=+1;
var+newArr+=+new+int[arr.length];
for(int+i+=+0;+i+<+arr.length;+i%2B%2B)
{
++newArr[i]+=+arr[arr.length+-+counter];
++counter%2B%2B;
}|code-block|syntax|javascript|794504|我没有测试它，但它应该在正确的轨道上。你有什么理由不想使用Array.Reverse吗？它可能是该算法的优化版本。|794505|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|$]]

try something like:

<pre><code>var counter = 1;
var newArr = new int[arr.length];
for(int i = 0; i &lt; arr.length; i++)
{
 newArr[i] = arr[arr.length - counter];
 counter++;
}
</code></pre>

I didn't test that but it should be on the right track. Any reason you dont want to use Array.Reverse? Its probably a well-optimized version of the algorithm.

blocks|key|4487811|text|在问题中替换//+some+code+here的代码是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4487812|for+(int+i+=+0;+i+<+arr.Length+/+2;+i%2B%2B)
{
+++int+tmp+=+arr[i];
+++arr[i]+=+arr[arr.Length+-+i+-+1];
+++arr[arr.Length+-+i+-+1]+=+tmp;
}|code-block|syntax|javascript|4487813|您应该只迭代数组的前半部分(arr.Length+/+2)。如果遍历整个数组(arr.Length)，它将颠倒两次，得到与开始之前相同的元素顺序。|4487814|entityMap^0|6|H|0|0|E|E|13|A|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

The code to be substituted in place of <code>// some code here</code> in the question is:

<pre><code>for (int i = 0; i &lt; arr.Length / 2; i++)
{
 int tmp = arr[i];
 arr[i] = arr[arr.Length - i - 1];
 arr[arr.Length - i - 1] = tmp;
}
</code></pre>

You should iterate only through the first half of the array (<code>arr.Length / 2</code>). If you iterate through the whole array (<code>arr.Length</code>), it will be reversed twice, yielding the same element order as before it started.

blocks|key|2412554|text|显然，你可以按相反的顺序复制到一个新的数组中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2412555|要进行“就地”操作，您可以从两端向中间操作:加载第一个和最后一个元素，然后将它们存储回去，第一个元素存储到最后一个位置，最后一个元素存储到第一个位置。然后做第二次和倒数第二次，等等。如果你有偶数个元素，你会进行N/2次迭代。如果是奇数，则执行(N-1)次/2迭代，并将中间元素保留在原来的位置。|2412556|在考虑缓存线大小和其他内存特征时，可能有其他算法会稍微快一些，但除非您处于真正的性能关键型情况下，否则它们不值得。|2412557|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Well, obviously you can just copy to a new array, in reverse order.

To do the operation "in place", you can work from both ends towards the middle: Load the first and last elements, then store them back, the first into the last location, and the last into the first location. Then do the second and the next-to-last, etc. If you have an even number of elements you do N/2 iterations. If an odd number you do (N-1)/2 iterations and leave the middle element where it was.

There are probably other algorithms that would be marginally faster when considering cache line size and other memory characteristics, but they wouldn't be worth it unless you were in a really performance-critical situation.

blocks|key|2696227|text|你可以用很多方法做到这一点，从最快的到最愚蠢的，比如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2696228|int[]+arr+=+new+int[]+{+1,2,3+};
arr+=+(from+a+in+arr+orderby+a+descending+select+a).ToArray();|code-block|syntax|javascript|2696229|但我不明白为什么你要追求如此徒劳的追求，如果这是为了给某人留下深刻印象，那么就使用这个而不是for循环:)|2696230|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|$]]

You can do this in many ways, from the most fast to the most stupid like:

<pre><code>int[] arr = new int[] { 1,2,3 };
arr = (from a in arr orderby a descending select a).ToArray();
</code></pre>

But I cannot understand why are you pursuing such a futile quest, if that is to impress someone somewhere then use this instead of the for loops :)

blocks|key|794607|text|Stack+stack=new+Stack;
var+newArr+=+new+int[arr.length];

for(int+i+=+0;+i+<+arr.length;+i%2B%2B)
{
++stack.push(arrr[i])
}
for(int+i+=+0;+i+<+arr.length;+i%2B%2B)
{
++newarr[i]=+stack.pop()
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|794608|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>Stack stack=new Stack;
var newArr = new int[arr.length];

for(int i = 0; i &lt; arr.length; i++)
{
 stack.push(arrr[i])
}
for(int i = 0; i &lt; arr.length; i++)
{
 newarr[i]= stack.pop()
}
</code></pre>

blocks|key|794632|text|int[]+array1+=+{+1,+2,+3,+4,+5+};

for+(int+x+=+4;+x+<+array1.Length+&&+x+!=+-1;+x--)
{
++++int+tmp;
++++tmp=array1[x];
++++Console.Write("{0}+",+tmp);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|794633|这就是我的解决方案。|unstyled|794634|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>int[] array1 = { 1, 2, 3, 4, 5 };

for (int x = 4; x &lt; array1.Length &amp;&amp; x != -1; x--)
{
 int tmp;
 tmp=array1[x];
 Console.Write("{0} ", tmp);
}
</code></pre>

That's my solution for this.

blocks|key|2412649|text|你可以向后循环：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2412650|int[]+arr=+new+int[]+{1,+2,+3,+4,+6};
for(int+i=arr.Length-1+;i>=+0+;+i--)
{
++++Console.WriteLine(arr[i].ToString());
}|code-block|syntax|javascript|2412651|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|$]]

You can just loop backwards:

<pre><code>int[] arr= new int[] {1, 2, 3, 4, 6};
for(int i=arr.Length-1 ;i&gt;= 0 ; i--)
{
 Console.WriteLine(arr[i].ToString());
}
</code></pre>

blocks|key|4487996|text|Console.WriteLine("Enter+a+string");
string+input+=+Console.ReadLine();

string+s+=+"";
for+(int+i+=+input.Length-1+;+i+>=+0;+i--)
{
++++s+=+s+%2B+input[i];
}
Console.WriteLine(s);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4487997|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>Console.WriteLine("Enter a string");
string input = Console.ReadLine();

string s = "";
for (int i = input.Length-1 ; i &gt;= 0; i--)
{
 s = s + input[i];
}
Console.WriteLine(s);
</code></pre>

blocks|key|794695|text|最好使用Array.Reverse方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794696|int[]+arr+={1,3,4,9,8};
Array.Reverse(arr);|code-block|syntax|javascript|794697|你可以阅读更多的描述Here|offset|length|794698|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/library/d3877932%2528v=vs.110%2529.aspx^0|0|0|A|4|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

It is better to use Array.Reverse method

<pre><code>int[] arr ={1,3,4,9,8};
Array.Reverse(arr);
</code></pre>

You can read more description <a href="https://msdn.microsoft.com/en-us/library/d3877932%28v=vs.110%29.aspx" rel="nofollow">Here</a>

blocks|key|794703|text|for+(int+i+=+0;+i+<+array.Length+-+i;+i%2B%2B)
+{
+++var+value+=+array[array.Length+-+i+-+1];
+++array[array.Length+-+i+-+1]+=+array[i];
+++array[i]+=+value;
+}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|794704|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>for (int i = 0; i &lt; array.Length - i; i++)
 {
 var value = array[array.Length - i - 1];
 array[array.Length - i - 1] = array[i];
 array[i] = value;
 }
</code></pre>

blocks|key|794738|text|//不使用反向方法，也不使用额外的数组//尝试从最后一个元素开始的yield运算符|type|unstyled|depth|inlineStyleRanges|entityRanges|data|794739|public+IEnumerable<int>+Reverse+(int[]+array)
{
++++for+(int+i+=+array.Length+-+1;+i+>=+0;+i--)+{
++++++++yield+return+array+[i];
++++}
}|code-block|syntax|javascript|794740|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|$]]

// without using Reverse method and without using additional array
// try yield operator starting from the last element

<pre><code>public IEnumerable&lt;int&gt; Reverse (int[] array)
{
 for (int i = array.Length - 1; i &gt;= 0; i--) {
 yield return array [i];
 }
}
</code></pre>

blocks|key|794754|text|int[]+triangles+=+new+int[]{0,1,2,3}++++
for+(int+j+=+triangles.Length;+j+>+(triangles.Length+/+2);+j--)
++++++++++++++++++++{
++++++++++++++++++++++++var+temp+=+triangles[j+-+1];
++++++++++++++++++++++++triangles[j+-+1]+=+triangles[triangles.Length+-+j];
++++++++++++++++++++++++triangles[triangles.Length+-+j]+=+temp;
++++++++++++++++++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|794755|我更喜欢从数组的末尾反转数组。我的解决方案是上面的。|unstyled|794756|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>int[] triangles = new int[]{0,1,2,3} 
for (int j = triangles.Length; j &gt; (triangles.Length / 2); j--)
 {
 var temp = triangles[j - 1];
 triangles[j - 1] = triangles[triangles.Length - j];
 triangles[triangles.Length - j] = temp;
 }
</code></pre>

I would prefer to reverse an array from the end of it. My solution's above.

blocks|key|4488117|text|可以使用单个for循环来做到这一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4488118|+++int[]+arr+={1,3,4,9,8};


+++for(int+i=arr.length-1;i>=0;i--)
+++{
++++++Console.Write(arr[i]);
+++}+++|code-block|syntax|javascript|4488119|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|$]]

Can do this with single for loop..
<pre><code> int[] arr ={1,3,4,9,8};


 for(int i=arr.length-1;i&gt;=0;i--)
 {
 Console.Write(arr[i]);
 } 
</code></pre>

blocks|key|2412858|text|基本上，您需要重新实现Array.Reverse(Array)。如果你看看implemented+in+the+framework本身，忽略了周围的许多技术细节，你会发现它只是在整个数组上调用它的三参数版本(反转数组的指定部分)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2412859|Array.Reverse(Array,Int32,Int32)是一个while循环，它交换元素并维护两个索引：|2412860|2412861|i指向反转部分的第一个元素，and|ordered-list-item|2412862|j指向反转部分的最后一个元素。|2412863|2412864|重写以替换问题中的//+some+code+here：|2412865|int+i+=+0;
int+j+=+arr.Length+-+1;
while+(i+<+j)
{
++++var+temp+=+arr[i];
++++arr[i]+=+arr[j];
++++arr[j]+=+temp;
++++i%2B%2B;
++++j--;
}|code-block|syntax|javascript|2412866|这比the+implementation+using+for-loop更容易掌握，做的算术更少，并且通过双重回归优雅地避开了陷阱。|2412867|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/library/d3877932.aspx|1|https://referencesource.microsoft.com/#mscorlib/system/array.cs,c1651950924e01b7|2|//stackoverflow.com/a/6088313/2157640^0|B|K|B|K|0|11|S|1|0|0|W|0|0|0|1|0|0|1|0|0|9|H|0|0|2|X|2|0^^$0|@$1|2|3|4|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@$9|1C|A|1D|1|1E]|$9|1F|A|1G|1|1H]]|E|$]]|$1|F|3|G|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|1M|8|@$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|L|3|M|5|K|7|1P|8|@$9|1Q|A|1R|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|1T|8|@$9|1U|A|1V|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|S|7|1W|8|@]|D|@]|E|$T|U]]|$1|V|3|W|5|6|7|1X|8|@]|D|@$9|1Y|A|1Z|1|20]]|E|$]]|$1|X|3|-4|5|6|7|21|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]|15|$5|10|11|12|E|$13|16]]|17|$5|10|11|12|E|$13|18]]]]

Basically, you are asked to reimplement <a href="https://msdn.microsoft.com/en-us/library/d3877932.aspx" rel="noreferrer"><code>Array.Reverse(Array)</code></a>. If you look at how it is <a href="https://referencesource.microsoft.com/#mscorlib/system/array.cs,c1651950924e01b7" rel="noreferrer">implemented in the framework</a> itself and ignore many technical details around, you’ll find that it just calls its three-parameter version (which reverses specified part of an array) on the whole array.

<code>Array.Reverse(Array,Int32,Int32)</code> is a while-loop that swaps elements and maintains two indexes:

<ol>
<li><code>i</code> points to the first element of the reversed part, and</li>
<li><code>j</code> points to the last element of the reversed part. </li>
</ol>

Rewritten to be substituted in place of <code>// some code here</code> in the question:

<pre><code>int i = 0;
int j = arr.Length - 1;
while (i &lt; j)
{
 var temp = arr[i];
 arr[i] = arr[j];
 arr[j] = temp;
 i++;
 j--;
}
</code></pre>

This is easier to grasp than <a href="//stackoverflow.com/a/6088313/2157640">the implementation using for-loop</a>, does less arithmetic and elegantly evades the gotcha with double reversion.

blocks|key|794862|text|+++function+printReverse(arr)+{
++++++for(var+i+=+arr.length+-+1;+i+>=+0;+i--){

++++console.log(arr[i]);
++++++}
+++}
printReverse([1,+2,+3,+6,+47,+88]);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|794863|​|unstyled|794864|function+printReverse(arr)+{
++++for+(var+i+=+arr.length+-+1;+i+>=+0;+i--)+{

++++++++console.log(arr[i]);
++++}
}
printReverse([1,+2,+3,+6,+47,+88])|794865|794866|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|N|8|@]|9|@]|A|$B|C]]|$1|I|3|E|5|F|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|F|7|P|8|@]|9|@]|A|$]]]|K|$]]

<pre><code> function printReverse(arr) {
 for(var i = arr.length - 1; i &gt;= 0; i--){

 console.log(arr[i]);
 }
 }
printReverse([1, 2, 3, 6, 47, 88]);
</code></pre>

<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>function printReverse(arr) {
 for (var i = arr.length - 1; i &gt;= 0; i--) {

 console.log(arr[i]);
 }
}
printReverse([1, 2, 3, 6, 47, 88])</code></pre>
</div>
</div>

blocks|key|2696541|text|char[]+strx+=+{+'1','2','3','4','5','6','7','8','9'+};
int+i+=+strx.Length;
string+ktr+="";

while+(i>0)
{
+++i--;
+++ktr+%2B=+strx[i];

+++if+(i==0)
+++{
++++++i+=+strx.Length;

++++++while+(i+>+0)
++++++{
+++++++++i--;
+++++++++strx[i]+=+ktr[i];
++++++}

+++}
}
int+j;
Console.WriteLine("Array+strx+in+reverse+order:+");
for+(j+=+0;+j+<+strx.Length;+j%2B%2B+)+
{
++Console.Write("{0}",+strx[j]);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2696542|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>char[] strx = { '1','2','3','4','5','6','7','8','9' };
int i = strx.Length;
string ktr ="";

while (i&gt;0)
{
 i--;
 ktr += strx[i];

 if (i==0)
 {
 i = strx.Length;

 while (i &gt; 0)
 {
 i--;
 strx[i] = ktr[i];
 }

 }
}
int j;
Console.WriteLine("Array strx in reverse order: ");
for (j = 0; j &lt; strx.Length; j++ ) 
{
 Console.Write("{0}", strx[j]);
}
</code></pre>

blocks|key|2412919|text|这是来自Array+legth等手表代码的如此简单的Start循环，您将理解：))|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2412920|++++++++int[]+arr+=+new+int[5]+{+1,+2,+3,+4,+5+};

++++++++for+(int+i+=+arr.Length-1;+i+>=+0;+i--)
++++++++{
++++++++++++Console.WriteLine(arr[i]);
++++++++}|code-block|syntax|javascript|2412921|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|$]]

That is So Simple Start loop from Array legth and so on watch code and you will get understand :))) 

<pre><code> int[] arr = new int[5] { 1, 2, 3, 4, 5 };

 for (int i = arr.Length-1; i &gt;= 0; i--)
 {
 Console.WriteLine(arr[i]);
 }
</code></pre>

blocks|key|2412943|text|我一点也不擅长循环。但在我看来这很简单-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2412944|+int[]+array1+=+{+1,+2,+3,+4,+5+};

+int[]+reverseArray+=+new+int[array1.Length];

+for+(int+i+=+0;+i+<=+array1.Length+-+1;+i%2B%2B)
++{
++++reverseArray[i]+=+array1[array1.Length+-+i+-+1];
++}|code-block|syntax|javascript|2412945|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|$]]

I am not good at loops at all. But this is what seems simple to me - 

<pre><code> int[] array1 = { 1, 2, 3, 4, 5 };

 int[] reverseArray = new int[array1.Length];

 for (int i = 0; i &lt;= array1.Length - 1; i++)
 {
 reverseArray[i] = array1[array1.Length - i - 1];
 }
</code></pre>

blocks|key|4488316|text|这是对任意关键点的datatype.Some数组进行反转的动态解决方案，算法中的关键点是先计算数组长度的一半，当数组索引具有相同的value.The阶段时，增加校验停止迭代，说明在这一阶段开始反转操作again.So，使用"goto语句“中断外循环。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|4488317|string[]+unreversed+=+{"A","B","C","D","E","F","G","H","I","J","K"};
++++++++++++int+q=unreversed.Length;
++++++++++++int+t+=+q+/+2;
++++++++++++var+temp1+=+"A";
++++++++++++for(int+i+=+0;i<unreversed.Length;i%2B%2B)
++++++++++++{
++++++++++++++++q+=+q+-+1;
++++++++++++++++for(int+k=q;k<=q;k%2B%2B)
++++++++++++++++{
++++++++++++++++++++if+(unreversed[k]+!=+unreversed[i]+&&+i!=t)
++++++++++++++++++++{
++++++++++++++++++++++++temp1+=+unreversed[i];
++++++++++++++++++++++++unreversed[i]+=+unreversed[k];
++++++++++++++++++++++++unreversed[k]+=+temp1;

++++++++++++++++++++}
++++++++++++++++++++else
++++++++++++++++++++{
++++++++++++++++++++++++goto+printarray;

++++++++++++++++++++}

++++++++++++++++}

++++++++++++}
++++++++++++printarray:
++++++++++++foreach+(var+k+in+unreversed)
++++++++++++{
++++++++++++++++Console.WriteLine(k);

++++++++++++}|code-block|syntax|javascript|4488318|entityMap^0|34|6|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|$]]

This is the dynamic solution for reversing the array of any datatype.Some of the key points in my algorithm is first calculate the half of array length and add check to stop iteration when array indexes have same value.The stage having same indexes depict that it start the reverse operation again.So at this stage break the outer loop by using "goto Statement".

<pre><code>string[] unreversed = {"A","B","C","D","E","F","G","H","I","J","K"};
 int q=unreversed.Length;
 int t = q / 2;
 var temp1 = "A";
 for(int i = 0;i&lt;unreversed.Length;i++)
 {
 q = q - 1;
 for(int k=q;k&lt;=q;k++)
 {
 if (unreversed[k] != unreversed[i] &amp;&amp; i!=t)
 {
 temp1 = unreversed[i];
 unreversed[i] = unreversed[k];
 unreversed[k] = temp1;

 }
 else
 {
 goto printarray;

 }

 }

 }
 printarray:
 foreach (var k in unreversed)
 {
 Console.WriteLine(k);

 }
</code></pre>

blocks|key|2413006|text|//Create+temp+array+with+the+same+size.
int[]+arrTemp+=+new+int[arr.Length];
int+i+=+0;
//Assign+last+value+of+arr+to+first+value+of+arrTemp
for+(int+j+=+arr.Length+-+1;+j+>=+0;+j--)
{
++++arrTemp[i]+=+arr[j];
++++i%2B%2B;
}
arr+=+arrTemp;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2413007|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>//Create temp array with the same size.
int[] arrTemp = new int[arr.Length];
int i = 0;
//Assign last value of arr to first value of arrTemp
for (int j = arr.Length - 1; j &gt;= 0; j--)
{
 arrTemp[i] = arr[j];
 i++;
}
arr = arrTemp;
</code></pre>

blocks|key|795067|text|我更喜欢使用索引的LINQ表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|795068|using+System.Linq;

int[]+arr+=+{+1,+3,+4,+9,+8+};
arr+=+arr.Select((n,+idx)+=>+new+{n,+idx})
++++.OrderByDescending(r+=>+r.idx)
++++.Select(r+=>+r.n).ToArray();|code-block|syntax|javascript|795069|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|$]]

I prefer a LINQ expression that uses an index:
<pre><code>using System.Linq;

int[] arr = { 1, 3, 4, 9, 8 };
arr = arr.Select((n, idx) =&gt; new {n, idx})
 .OrderByDescending(r =&gt; r.idx)
 .Select(r =&gt; r.n).ToArray();
</code></pre>

blocks|key|795071|text|++++public+int[]+Reverse(params+int[]+numbers)
++++{
++++++++for+(int+i+=+0;+i+<+numbers.Length+/+2;+i%2B%2B)
++++++++{
++++++++++++int+tmp+=+numbers[i];
++++++++++++numbers[i]+=+numbers[numbers.Length+-+i+-+1];
++++++++++++numbers[numbers.Length+-+i+-+1]+=+tmp;
++++++++}

++++++++return+numbers;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|795072|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> public int[] Reverse(params int[] numbers)
 {
 for (int i = 0; i &lt; numbers.Length / 2; i++)
 {
 int tmp = numbers[i];
 numbers[i] = numbers[numbers.Length - i - 1];
 numbers[numbers.Length - i - 1] = tmp;
 }

 return numbers;
 }
</code></pre>

blocks|key|795098|text|下面是一个使用Length()函数和一个简单的for循环来反转数组的示例。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|795099|using+System;
using+System.Collections.Generic;
using+System.Linq;
using+System.Text.RegularExpressions;

namespace+Rextester
{
++++public+class+Program
++++{
++++++++public+static+void+Main(string[]+args)
++++++++{
++++++++++++int[]+arr+=+new+int[]+{4,+8,+2,+9,+5,+5};
++++++++++++
++++++++++++int+length+=+arr.Length;
++++++++++++
++++++++++++for(int+i+=+0;+i+<+arr.Length;+i%2B%2B)
++++++++++++{
+++++++++++++++Console.WriteLine(arr[length-1]+%2B+"+");
+++++++++++++++length+=+length+-+1;
++++++++++++}++++++++++++
++++++++}++++++++
++++}
}|code-block|syntax|javascript|795100|entityMap^0|7|8|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|$]]

Here is an example of reversing an array using the <code>Length()</code> function and a simple for loop.
<pre><code>using System;
using System.Collections.Generic;
using System.Linq;
using System.Text.RegularExpressions;

namespace Rextester
{
 public class Program
 {
 public static void Main(string[] args)
 {
 int[] arr = new int[] {4, 8, 2, 9, 5, 5};
 
 int length = arr.Length;
 
 for(int i = 0; i &lt; arr.Length; i++)
 {
 Console.WriteLine(arr[length-1] + &quot; &quot;);
 length = length - 1;
 } 
 } 
 }
}
</code></pre>

How to reverse an array (in C#) without using <code>Array.Reverse()</code> method?

For example,

<pre><code>int[] arr = {1,3,4,9,8};
// some code here
Console.WriteLine(string.Join(",", arr));
</code></pre>

should result in

<pre><code>8,9,4,3,1
</code></pre>

I got this as an interview task.

Reverse an array without using Array.Reverse()

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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如何在不使用Array.Reverse()方法的情况下反转数组(在C#中)？例如,int[] arr = {1,3,4,9,8};// some code hereConsole.WriteLine(string.Join(",", arr));应该会导致8,9,4,3,1这是我的面试任务。

问不使用Array.Reverse()反转数组
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问不使用Array.Reverse()反转数组EN