blocks|key|5807907|text|假设@DominicTobias是正确的，你可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5807908|​|5807909|console.log(+
++++%2Bprompt("Enter+a+value").split("").reverse().join("")+
)|code-block|syntax|javascript|5807910|5807911|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>Assuming @DominicTobias is correct, you can use this:</p>

<p><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>console.log( 
    +prompt("Enter a value").split("").reverse().join("") 
)</code></pre>
</div>
</div>
</p>


blocks|key|5807981|text|首先，我认为你没有使用数组来存储数字。您正在使用java脚本变量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5807982|试试这段代码，看看它是否能正常工作。|5807983|var+a+=+prompt("Enter+a+value");
var+z+=+a;
var+reverse+=+0;
while(z+>+0)
{
++++var+digit+=+z+%25+10;
++++reverse+=+(reverse+*+10)+%2B+digit;
++++z+=+parseInt(z+/+10);
}
alert("reverse+=+"+%2B+reverse);|code-block|syntax|javascript|5807984|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|$]]

<p>Firstly, I don't think you are using an array to store the number. You are using a java script variable.</p>

<p>Try out this code and see if it works.</p>

<pre><code>var a = prompt("Enter a value");
var z = a;
var reverse = 0;
while(z &gt; 0)
{
    var digit = z % 10;
    reverse = (reverse * 10) + digit;
    z = parseInt(z / 10);
}
alert("reverse = " + reverse);
</code></pre>


blocks|key|2279751|text|解释|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2279752|使用JavaScript+reverse()数组方法可以颠倒数组元素的顺序。|offset|length|style|CODE|2279753|代码|2279754|​|2279755|var+a+=+prompt("Enter+a+value");
var+arr+=+[];

for+(var+i+=+0;+i+<+a.length;+i%2B%2B)+{
++++arr[i]+=+a.charAt(i);
}
arr.reverse();
alert(arr);|code-block|syntax|javascript|2279756|2279757|entityMap^0|0|D|9|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@$D|V|E|W|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Y|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|Z|8|@]|9|@]|A|$O|P]]|$1|Q|3|K|5|6|7|10|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|S|$]]

<h2>Explanation</h2>
<p>Using the JavaScript <code>reverse()</code> array method you can reverse the order of the array elements.</p>
<h2>Code</h2>
<p><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>var a = prompt("Enter a value");
var arr = [];

for (var i = 0; i &lt; a.length; i++) {
    arr[i] = a.charAt(i);
}
arr.reverse();
alert(arr);</code></pre>
</div>
</div>
</p>


blocks|key|5808048|text|我的反转字符串的解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5808049|var+text+=+""
var+i+=+0
var+array+=+["1",+"2",+"3"]
var+number+=+array.length
var+arrayFinal+=+[]

for+(i+=+0;+i+<+array.length;+i%2B%2B)+{
++++text+=+array[number+-+1]
++++arrayFinal.push(text)
++++text+=+""
++++number+=+number+-+1
}

console.log(arrayFinal)|code-block|syntax|javascript|5808050|希望这能有所帮助！|5808051|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|$]]

<p>My solution to reverse a string:</p>

<pre><code>var text = ""
var i = 0
var array = ["1", "2", "3"]
var number = array.length
var arrayFinal = []

for (i = 0; i &lt; array.length; i++) {
    text = array[number - 1]
    arrayFinal.push(text)
    text = ""
    number = number - 1
}

console.log(arrayFinal)
</code></pre>

<p>Hopes This Helps!</p>


blocks|key|1758005|text|或者，作为一行程序(+x包含要反转的整数)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1758006|revX=x.toFixed(0).split('').reverse().join('')-0;|code-block|syntax|javascript|1758007|数字将被分成单独的数字，颠倒过来，然后重新组合成一个字符串。然后，-0再次将其转换为数字。|1758008|entityMap^0|B|1|0|0|X|2|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|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

<p>Or, as a one-liner ( <code>x</code> contains the integer number to be inversed):</p>

<pre><code>revX=x.toFixed(0).split('').reverse().join('')-0;
</code></pre>

<p>The number will be separated into its individual digits, reversed and then reassembled again into a string. The <code>-0</code> then converts it into a number again.</p>


blocks|key|2279960|text|低级整数反转：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2279961|​|2279962|function+flipInt(n){
++++var+digit,+result+=+0

++++while(+n+){
++++++++digit+=+n+%25+10++//++Get+right-most+digit.+Ex.+123/10+→+12.3+→+3
++++++++result+=+(result+*+10)+%2B+digit++//++Ex.+123+→+1230+%2B+4+→+1234
++++++++n+=+n/10%7C0++//++Remove+right-most+digit.+Ex.+123+→+12.3+→+12
++++}++
++
++++return+result
}


//+Usage:+
alert(
++"Reversed+number:+"+%2B+flipInt(+%2Bprompt("Enter+a+value")+)+
)|code-block|syntax|javascript|2279963|2279964|上面的代码使用bitwise+operators进行快速计算|offset|length|2279965|2279966|此方法比其他方法快得多，其他方法将数字转换为数组，然后将其反转并再次联接。这是一个低级的快速解决方案。|blockquote|2279967|2279968|2279969|插图表格：|2279970|2279971|const+delay+=+(ms+=+1000)+=>+new+Promise(res+=>+setTimeout(res,+ms))
const+table+=+document.querySelector('tbody')

async+function+printLine(s1,+s2,+op){
++table.innerHTML+%2B=+`<tr>
++++<td>${s1}</td>
++++<td>${s2%7C%7C''}</td>
++</tr>`
}

async+function+steps(){
++printLine(123)
++await+delay()
++
++printLine('12.3+→')
++await+delay()
++
++printLine(12,+3)
++await+delay()
++
++printLine('1.2',+'3+&times;+10')
++await+delay()
++
++printLine('1.2+→',+30)
++await+delay()
++
++printLine(1,+32)
++await+delay()
++
++printLine(1,+'32+&times;+10')
++await+delay()
++
++printLine('1+→',+320)
++await+delay()
++
++printLine('',+321)
++await+delay()
}

steps()|2279972|table{+width:+200px;+}
td+{
++border:+1px+dotted+#999;
}|2279973|<table>
++<thead>
++++<tr>
++++++<th>Current</th>
++++++<th>Output</th>
++++</tr>
++</thead>
++<tbody>
++</tbody>
</table>|2279974|2279975|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators^0|0|0|0|0|7|H|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1C|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1D|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|1E|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1F|8|@]|9|@$L|1G|M|1H|1|1I]]|A|$]]|$1|N|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1K|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1N|8|@]|9|@]|A|$]]|$1|V|3|C|5|6|7|1O|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1P|8|@]|9|@]|A|$G|H]]|$1|Y|3|Z|5|F|7|1Q|8|@]|9|@]|A|$G|H]]|$1|10|3|11|5|F|7|1R|8|@]|9|@]|A|$G|H]]|$1|12|3|C|5|6|7|1S|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|14|$15|$5|16|17|18|A|$19|1A]]]]

<h2>Low-level integer numbers reversing:</h2>
<p><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 flipInt(n){
    var digit, result = 0

    while( n ){
        digit = n % 10  //  Get right-most digit. Ex. 123/10 → 12.3 → 3
        result = (result * 10) + digit  //  Ex. 123 → 1230 + 4 → 1234
        n = n/10|0  //  Remove right-most digit. Ex. 123 → 12.3 → 12
    }  
  
    return result
}


// Usage: 
alert(
  "Reversed number: " + flipInt( +prompt("Enter a value") ) 
)</code></pre>
</div>
</div>
</p>
<p>The above code uses <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators" rel="noreferrer">bitwise operators</a> for quick math</p>
<blockquote>
<p>This method is <em><strong>MUCH FASTER</strong></em> than other methods which convert the number to an Array and then reverse it and join it again. This is a low-level blazing-fast solution.</p>
</blockquote>
<h3>Illustration table:</h3>
<p><div class="snippet" data-lang="js" data-hide="true" data-console="true" data-babel="false">
<div class="snippet-code snippet-currently-hidden">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const delay = (ms = 1000) =&gt; new Promise(res =&gt; setTimeout(res, ms))
const table = document.querySelector('tbody')

async function printLine(s1, s2, op){
  table.innerHTML += `&lt;tr&gt;
    &lt;td&gt;${s1}&lt;/td&gt;
    &lt;td&gt;${s2||''}&lt;/td&gt;
  &lt;/tr&gt;`
}

async function steps(){
  printLine(123)
  await delay()
  
  printLine('12.3 →')
  await delay()
  
  printLine(12, 3)
  await delay()
  
  printLine('1.2', '3 &amp;times; 10')
  await delay()
  
  printLine('1.2 →', 30)
  await delay()
  
  printLine(1, 32)
  await delay()
  
  printLine(1, '32 &amp;times; 10')
  await delay()
  
  printLine('1 →', 320)
  await delay()
  
  printLine('', 321)
  await delay()
}

steps()</code></pre>
<pre class="snippet-code-css lang-css prettyprint-override"><code>table{ width: 200px; }
td {
  border: 1px dotted #999;
}</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;table&gt;
  &lt;thead&gt;
    &lt;tr&gt;
      &lt;th&gt;Current&lt;/th&gt;
      &lt;th&gt;Output&lt;/th&gt;
    &lt;/tr&gt;
  &lt;/thead&gt;
  &lt;tbody&gt;
  &lt;/tbody&gt;
&lt;/table&gt;</code></pre>
</div>
</div>
</p>


blocks|key|2279977|text|假设您可能希望将其反转为一个真实的数字，而不是一个字符串，请尝试以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2279978|function+reverseNumber(num){
++++num+=+num+%2B+'';
++++++++let+reversedText+=+num.split('').reverse().join('');
++++++++let+reversedNumber+=+parseInt(reversedText,+10);
++++console.log("reversed+number:+",+reversedNumber);
++++++++return+reversedNumber;
++}|code-block|syntax|javascript|2279979|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|$]]

<p>Assuming you may want to reverse it as a true number and not a string try the following:</p>

<pre><code>function reverseNumber(num){
    num = num + '';
        let reversedText = num.split('').reverse().join('');
        let reversedNumber = parseInt(reversedText, 10);
    console.log("reversed number: ", reversedNumber);
        return reversedNumber;
  }
</code></pre>


blocks|key|5808146|text|使用JavaScript+reverse()和Math.sign()可以反转正数和负数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5808147|var+enteredNum+=+prompt("Enter+integer");

function+reverseInteger(enteredNum)+{
++++const+reveredNumber+=+enteredNum.toString().split('').reverse().join('');
++++return+parseInt(reveredNumber)*Math.sign(enteredNum);
}

alert(reverseInteger(enteredNum));|code-block|syntax|javascript|5808148|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|$]]

<p>Using JavaScript reverse() and Math.sign() you can reverse a number both positive and negative numbers.</p>

<pre><code>var enteredNum = prompt("Enter integer");

function reverseInteger(enteredNum) {
    const reveredNumber = enteredNum.toString().split('').reverse().join('');
    return parseInt(reveredNumber)*Math.sign(enteredNum);
}

alert(reverseInteger(enteredNum));
</code></pre>


blocks|key|5808204|text|最近有人问我如何解决这个问题，这是我最初的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5808205|5808206|5808207|+所需的输出：+123+=>+321，-15+=>+-51,500+=>+5|blockquote|offset|length|style|BOLD|5808208|5808209|5808210|​|5808211|function+revInt(num)+{
//+Use+toString()+to+convert+it+into+a+String
//+Use+the+split()+method+to+return+a+new+array:+-123+=>+['-',+'1','2','3']
//+Use+the+reverse()+method+to+reverse+the+new+created+array:+['-',+'1','2','3']+=>+['3','2','1','-'];
//+Use+the+join()+method+to+join+all+elements+of+the+array+into+a+string
++let+val+=+num.toString().split('').reverse().join('');
++++//+If+the+entered+number+was+negative,+then+that+'-'+would+be+the+last+character+in
+++//++our+newly+created+String,+but+we+don't+want+that,+instead+what+we+want+is
++//++for+it+to+be+the+first+one.+So,+this+was+the+solution+from+the+top+of+my+head.

//+The+endsWith()+method+determines+whether+a+string+ends+with+the+characters+of+a+specified+string
++if+(val.endsWith('-'))+{
++++val+=+'-'+%2B+val;
++++++return+parseInt(val);
++}
++++++return+parseInt(val);
}

console.log(revInt(-123));|code-block|syntax|javascript|5808212|5808213|5808214|更好的解决方案：|5808215|在我进一步思考之后，我想出了以下几点：|5808216|5808217|+++//+Here+we're+converting+the+result+of+the+same+functions+used+in+the+above+example+to+
//+an+Integer+and+multiplying+it+by+the+value+returned+from+the+Math.sign()+function.

//+NOTE:+The+Math.sign()+function+returns+either+a+positive+or+negative+%2B/-+1,+
//+indicating+the+sign+of+a+number+passed+into+the+argument.+++

function+reverseInt(n)+{
++++++return+parseInt(n.toString().split('').reverse().join(''))+*+Math.sign(n)
}

console.log(reverseInt(-123));|5808218|5808219|注意:第二种解决方案要简单得多，我想|5808220|entityMap^0|0|0|0|1|6|0|0|0|0|0|0|0|0|8|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|C|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1A|8|@$G|1B|H|1C|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1F|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1G|8|@]|9|@]|A|$R|S]]|$1|T|3|N|5|6|7|1H|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1J|8|@$G|1K|H|1L|I|J]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1M|8|@]|9|@]|A|$]]|$1|Z|3|N|5|6|7|1N|8|@]|9|@]|A|$]]|$1|10|3|11|5|Q|7|1O|8|@]|9|@]|A|$R|S]]|$1|12|3|N|5|6|7|1P|8|@]|9|@]|A|$]]|$1|13|3|14|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|15|3|-4|5|6|7|1R|8|@]|9|@]|A|$]]]|16|$]]

<p>I was recently asked how to solve this problem and this was my initial solution:  </p>

<hr>

<blockquote>
  <p><strong><em>The desired output:</em></strong> <em>123 => 321, -15 => -51, 500 => 5</em></p>
</blockquote>

<p><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 revInt(num) {
// Use toString() to convert it into a String
// Use the split() method to return a new array: -123 =&gt; ['-', '1','2','3']
// Use the reverse() method to reverse the new created array: ['-', '1','2','3'] =&gt; ['3','2','1','-'];
// Use the join() method to join all elements of the array into a string
  let val = num.toString().split('').reverse().join('');
    // If the entered number was negative, then that '-' would be the last character in
   //  our newly created String, but we don't want that, instead what we want is
  //  for it to be the first one. So, this was the solution from the top of my head.

// The endsWith() method determines whether a string ends with the characters of a specified string
  if (val.endsWith('-')) {
    val = '-' + val;
      return parseInt(val);
  }
      return parseInt(val);
}

console.log(revInt(-123));</code></pre>
</div>
</div>
</p>

<hr>

<p><strong><em>A way better solution:</em></strong></p>

<p>After I gave it some more thought, I came up with the following: </p>

<p><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>   // Here we're converting the result of the same functions used in the above example to 
// an Integer and multiplying it by the value returned from the Math.sign() function.

// NOTE: The Math.sign() function returns either a positive or negative +/- 1, 
// indicating the sign of a number passed into the argument.   

function reverseInt(n) {
      return parseInt(n.toString().split('').reverse().join('')) * Math.sign(n)
}

console.log(reverseInt(-123));</code></pre>
</div>
</div>
</p>

<p><em>NOTE: The 2nd solution is much more straightforward, IMHO</em> </p>


blocks|key|5808238|text|在论文中应用倒数逻辑并尝试，你必须关心除法，因为它给出了浮点值，这就是为什么我们必须使用parseInt()。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5808239|<head>
<script>
++++function+palindrome()
++++{
++++++++var+a+=+document.getElementById('str').value;
++++++++var+r=0+,t=0;
++++++++while(a>0){
++++++++++++r=a%2510;
++++++++++++t=t*10%2Br;
++++++++++++a=parseInt(a/10);
++++++++}
++++++++
++++++++document.write(t);
++++++++
++++}
++++</script>

</head>

<body>
<form>
++++<input+type="text"+id="str"/>
++++<input+type="submit"+onClick="palindrome()"+/>
<form>
</body>
</html>|code-block|syntax|javascript|5808240|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|$]]

<p>Apply Logic of reversing number in paper and try,and you have to care about dividing because it gives float values thats why we have to use parseInt().</p>
<pre><code>&lt;head&gt;
&lt;script&gt;
    function palindrome()
    {
        var a = document.getElementById('str').value;
        var r=0 ,t=0;
        while(a&gt;0){
            r=a%10;
            t=t*10+r;
            a=parseInt(a/10);
        }
        
        document.write(t);
        
    }
    &lt;/script&gt;

&lt;/head&gt;

&lt;body&gt;
&lt;form&gt;
    &lt;input type=&quot;text&quot; id=&quot;str&quot;/&gt;
    &lt;input type=&quot;submit&quot; onClick=&quot;palindrome()&quot; /&gt;
&lt;form&gt;
&lt;/body&gt;
&lt;/html&gt;
</code></pre>


blocks|key|1758126|text|function+add(+num:number){++//159
let+d+:+number;+
let+a+:+number+=0;++
while(num+>+0){++++//159++15++1
++++d+=+num+%25+10;++
++++a+=+a+*+10+%2B+d;++//9++95+++951
++++num+=+Math.floor(num/10);++//+15++1++0
}
+return+a;++//951
}
console.log(add(159));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1758127|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>function add( num:number){  //159
let d : number; 
let a : number =0;  
while(num &gt; 0){    //159  15  1
    d = num % 10;  
    a = a * 10 + d;  //9  95   951
    num = Math.floor(num/10);  // 15  1  0
}
 return a;  //951
}
console.log(add(159));
</code></pre>


blocks|key|1761017|text|甜蜜而简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1761018|function+reverseNumber(num){
+++return+parseInt(num.toString().split("").reverse().join(""));
}|code-block|syntax|javascript|1761019|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|$]]

<p>Sweet and simple:</p>
<pre><code>function reverseNumber(num){
   return parseInt(num.toString().split(&quot;&quot;).reverse().join(&quot;&quot;));
}
</code></pre>


blocks|key|5808319|text|上面的代码对负数不起作用。相反，请使用以下内容：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5808320|/**
+*+@param+{number}+x
+*+@return+{boolean}
+*/
var+isPalindrome+=+function(x)+{
++++return++((x>=0)+?+((x==(x+=+parseInt(x.toString().split("").reverse().join(""))))+?+true:false)+:+false);
};|code-block|syntax|javascript|5808321|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|$]]

<p>The above code will not work for negative numbers. Instead, use the following:</p>
<pre><code>/**
 * @param {number} x
 * @return {boolean}
 */
var isPalindrome = function(x) {
    return  ((x&gt;=0) ? ((x==(x = parseInt(x.toString().split(&quot;&quot;).reverse().join(&quot;&quot;)))) ? true:false) : false);
};
</code></pre>


blocks|key|1758180|text|这是我的解决方案，没有预定义函数的纯JS。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1758181|++++function+reverseNum(number)+{
++var+result+=+0,
++++counter+=+0;
++for+(i+=+number;+i+>=+1;+i+=+i+/+10+-+(i+%25+10)+*+0.1)+{
++++counter+=+i+%25+10;
++++result+=+result+*+10+%2B+counter;
++}
++return+result;
}

console.log(reverseNum(547793));|code-block|syntax|javascript|1758182|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|$]]

<p>This is my solution, pure JS without predefined functions.</p>
<pre><code>    function reverseNum(number) {
  var result = 0,
    counter = 0;
  for (i = number; i &gt;= 1; i = i / 10 - (i % 10) * 0.1) {
    counter = i % 10;
    result = result * 10 + counter;
  }
  return result;
}

console.log(reverseNum(547793));
</code></pre>


blocks|key|2280149|text|反转一个数字，而不使用递归方法将其转换为字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2280150|​|2280151|const+num+=+4578;

const+by10+=+(num)+=>+{
++return+Math.floor(num+/+10);
};
const+remBy10+=+(num)+=>+{
++return+Math.floor(num+%25+10);
};

const+reverseNum+=+(num,+str+=+"")+=>+{
++if+(num.toString().length+==+1)+return+(str+%2B=+num);
++return+reverseNum(by10(num),+(str+%2B=+remBy10(num)));
};

console.log(reverseNum(num,+""));|code-block|syntax|javascript|2280152|2280153|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>Reversing a number without converting it into the string using the recursive approach.</p>
<p><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>const num = 4578;

const by10 = (num) =&gt; {
  return Math.floor(num / 10);
};
const remBy10 = (num) =&gt; {
  return Math.floor(num % 10);
};

const reverseNum = (num, str = "") =&gt; {
  if (num.toString().length == 1) return (str += num);
  return reverseNum(by10(num), (str += remBy10(num)));
};

console.log(reverseNum(num, ""));</code></pre>
</div>
</div>
</p>


blocks|key|1758188|text|var+reverse+=+function(x)+{
+++if+(x+>+2147483647+%7C%7C+x+<+-2147483648+%7C%7C+x+===+0)+{
+++++++return+0;
+++}

+++let+isNegative+=+false;

+++if(x+<+0){
+++++++isNegative+=+true;
+++++++x+=+-x;
+++}

+++const+length+=+parseInt(Math.log10(x));
+++let+final+=+0;
+++let+digit+=+x;
+++let+mul+=+0;

+++for(let+i+=+length+;+i+>=+0;+i--){
+++++++digit+=+parseInt(x+/+(10**i));
+++++++mul+=+10**(length-i);
+++++++final+=+final+%2B+digit+*+mul;+
+++++++x+=+parseInt(x+%25+10**i);
+++}

+++if+(final+>+2147483647+%7C%7C+final+<+-2147483648+)+{
+++++++return+0;
+++}

+++if(isNegative){
+++++++return+-final;
+++}
+++else{
+++++++return+final;
+++}
};

+++
console.log(reverse(1534236469));
console.log(reverse(-123));
console.log(reverse(120));
console.log(reverse(0));
console.log(reverse(2,147,483,648));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1758189|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-js prettyprint-override"><code>var reverse = function(x) {
   if (x &gt; 2147483647 || x &lt; -2147483648 || x === 0) {
       return 0;
   }

   let isNegative = false;

   if(x &lt; 0){
       isNegative = true;
       x = -x;
   }

   const length = parseInt(Math.log10(x));
   let final = 0;
   let digit = x;
   let mul = 0;

   for(let i = length ; i &gt;= 0; i--){
       digit = parseInt(x / (10**i));
       mul = 10**(length-i);
       final = final + digit * mul; 
       x = parseInt(x % 10**i);
   }

   if (final &gt; 2147483647 || final &lt; -2147483648 ) {
       return 0;
   }

   if(isNegative){
       return -final;
   }
   else{
       return final;
   }
};

   
console.log(reverse(1534236469));
console.log(reverse(-123));
console.log(reverse(120));
console.log(reverse(0));
console.log(reverse(2,147,483,648));
   
</code></pre>


blocks|key|2280197|text|最简单的解决方案是反转js中的任何整数。不适用于浮点数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2280198|const+i2a+=+number.toString().split("");
const+a2i+=+parseInt(i2a.reverse().join(""));
console.log(a2i);|code-block|syntax|javascript|2280199|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|$]]

<p>The simplest solution is to reverse any integer in js. Doesn't work with float.</p>
<pre><code>const i2a = number.toString().split(&quot;&quot;);
const a2i = parseInt(i2a.reverse().join(&quot;&quot;));
console.log(a2i);
</code></pre>


blocks|key|5808434|text|最简单的方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5808435|5808436|将其隐藏为字符串，然后应用reverse()方法|ordered-list-item|offset|length|style|CODE|5808437|将其改回number|5808438|检查所提供的值是否为负数或正数|5808439|5808440|下面是我对此的解决方案。|5808441|​|5808442|function+reverseInt(n)+{
++const+reversed+=
++++n.toString().split('').reverse().join('');
++return+parseInt(reversed)+*+Math.sign(n);
}

console.log(reverseInt(12345));|code-block|syntax|javascript|5808443|5808444|entityMap^0|0|0|D|9|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|12|8|@$F|13|G|14|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|E|7|16|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@]|9|@]|A|$]]|$1|S|3|T|5|U|7|1A|8|@]|9|@]|A|$V|W]]|$1|X|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|Z|$]]

<p>The simplest way is to</p>
<ol>
<li>Covert it into a string and apply the  <code>reverse()</code> method</li>
<li>Change it back to number</li>
<li>Check for the value provided if negative or positive with <strong>Math.sign()</strong></li>
</ol>
<p>Below is my solution to that.</p>
<p><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 reverseInt(n) {
  const reversed =
    n.toString().split('').reverse().join('');
  return parseInt(reversed) * Math.sign(n);
}

console.log(reverseInt(12345));</code></pre>
</div>
</div>
</p>


<p>Below is my source code to reverse (as in a mirror) the given number.
I need to reverse the number using the reverse method of arrays.</p>
<pre><code>var a = prompt(&quot;Enter a value&quot;);
var b, sum = 0;
var z = a;
while(a &gt; 0)
{
  b = a % 10;
  sum = sum * 10 + b;
  a = parseInt(a / 10);
}
alert(sum);
</code></pre>


JavaScript: How to reverse a number?

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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下面是我用来反转(就像在镜像中)给定数字的源代码。我需要使用数组的反转方法来反转数字。var a = prompt("Enter a value");var b, sum = 0;var z = a;while(a > 0){  b = a % 10;  sum = sum * 10 + b;  a = parseIn...

问JavaScript:如何反转一个数字？
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问JavaScript:如何反转一个数字？EN