blocks|key|1842165|text|您可以首先使用reduce获取一个对象，并对每个number元素进行计数，然后在Object.keys上使用filter返回非重复数字数组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1842166|​|1842167|var+arr=[-1,2,5,6,2,9,-1,6,5,-1,3];
var+obj+=+arr.reduce((r,+e)+=>+(r[e]+=+(r[e]+%7C%7C+0)+%2B+1,+r),+{});
var+uniq+=+Object.keys(obj).filter(e+=>+obj[e]+==+1).map(Number)

console.log(uniq)|code-block|syntax|javascript|1842168|1842169|有for循环的解决方案。|1842170|1842171|var+arr+=+[-1,+2,+5,+6,+2,+9,+-1,+6,+5,+-1,+3];

var+uniq+=+[];
for+(var+i+=+0;+i+<+arr.length;+i%2B%2B)+{
++for+(var+j+=+0;+j+<+arr.length;+j%2B%2B)+{
++++if+(arr[i]+==+arr[j]+&&+i+!=+j)+break;
++++else+if+(j+==+arr.length+-+1)+uniq.push(arr[i])
++}
}

console.log(uniq)|1842172|1842173|entityMap^0|7|6|14|B|1I|6|0|0|0|0|1|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|14|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|P|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|S|3|G|5|6|7|1A|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|U|$]]

<p>You could first use <code>reduce</code> to get one object with count for each number element and then <code>filter</code> on <code>Object.keys</code> to return array of non-repeating numbers.</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>var arr=[-1,2,5,6,2,9,-1,6,5,-1,3];
var obj = arr.reduce((r, e) =&gt; (r[e] = (r[e] || 0) + 1, r), {});
var uniq = Object.keys(obj).filter(e =&gt; obj[e] == 1).map(Number)

console.log(uniq)</code></pre>
</div>
</div>
</p>

<p>Solution with <code>for</code> loop.</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>var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];

var uniq = [];
for (var i = 0; i &lt; arr.length; i++) {
  for (var j = 0; j &lt; arr.length; j++) {
    if (arr[i] == arr[j] &amp;&amp; i != j) break;
    else if (j == arr.length - 1) uniq.push(arr[i])
  }
}

console.log(uniq)</code></pre>
</div>
</div>
</p>


blocks|key|1967564|text|只根据OP请求过滤唯一元素：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1967565|这将根据请求使用循环。它返回一个只包含在原始数组中出现一次的元素的数组。|1967566|​|1967567|var+arr+=+[-1,+2,+5,+6,+2,+9,+-1,+6,+5,+-1,+3];
var+n+=+arr.length;
var+result+=+[];

function+nonrep()+{
++for+(var+i+=+0;+i+<+n;+i%2B%2B)+{
++++for+(var+j=0+;+j+<+n;+j%2B%2B)
++++++if+(i!=j+&&+arr[i]==arr[j])
++++++++break;
++++if(j==n)
++++++++result.push(arr[i]);
++}
++return+result;
}
console.log(nonrep())|code-block|syntax|javascript|1967568|1967569|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|Q|8|@]|9|@]|A|$I|J]]|$1|K|3|E|5|6|7|R|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|M|$]]

<h1>Filtering only unique elements according to OP request:</h1>
<p>This uses for loops, as requested. It returns an array containing only elements appearing once in the original array.</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>var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
var n = arr.length;
var result = [];

function nonrep() {
  for (var i = 0; i &lt; n; i++) {
    for (var j=0 ; j &lt; n; j++)
      if (i!=j &amp;&amp; arr[i]==arr[j])
        break;
    if(j==n)
        result.push(arr[i]);
  }
  return result;
}
console.log(nonrep())</code></pre>
</div>
</div>
</p>


blocks|key|1967578|text|为此，您可以使用过滤器和indexOf：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1967579|console.log(
++[-1,+2,+5,+6,+2,+9,+-1,+6,+5,+-1,+3].filter((v,+i,+a)+=>+a.indexOf(v,+i+%2B+1)+===+-1+)
);|code-block|syntax|javascript|1967580|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf^0|8|3|0|C|7|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

<p>You can use <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter" rel="nofollow noreferrer">filter</a> and <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf" rel="nofollow noreferrer">indexOf</a> for that:</p>

<pre><code>console.log(
  [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3].filter((v, i, a) =&gt; a.indexOf(v, i + 1) === -1 )
);
</code></pre>


blocks|key|1842263|text|一些变化：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1842264|将所有变量声明移到函数中。|unordered-list-item|1842265|对提交的数组使用函数参数，保持函数的纯正。|1842266|预先在函数顶部声明所有需要的变量。|1842267|以数组作为结果数组unique。|offset|length|style|CODE|1842268|检查i和j，如果等于continue+(内)循环。|1842269|检查i和j的值并退出(内部)循环，因为找到了一个重复的循环。|1842270|在内循环结束时进行检查，检查索引j和数组l的长度，如果相同，则将值推送到unique。|1842271|在外部循环的末尾使用一个带有return数组的unique语句。|1842272|​|1842273|function+getUnique(array)+{
++++var+l+=+array.length,
++++++++i,+j,
++++++++unique+=+[];

++++for+(i+=+0;+i+<+l;+i%2B%2B)+{
++++++++for+(j+=+0;+j+<+l;+j%2B%2B)+{
++++++++++++if+(i+===+j)+{
++++++++++++++++continue;
++++++++++++}
++++++++++++if+(array[i]+===+array[j])+{
++++++++++++++++break;
++++++++++++}
++++++++}
++++++++if+(j+===+l)+{
++++++++++++unique.push(array[i]);
++++++++}
++++}
++++return+unique;
}

console.log(getUnique([-1,+2,+5,+6,+2,+9,+-1,+6,+5,+-1,+3]));|code-block|syntax|javascript|1842274|1842275|另一种解决方案可能是检查indexOf和lastIndexOf是否返回相同的值。然后你找到了一个独特的价值。|1842276|1842277|var+array+=+[-1,+2,+5,+6,+2,+9,+-1,+6,+5,+-1,+3],
++++unique+=+array.filter((v,+i)+=>+array.indexOf(v)+===+array.lastIndexOf(v));

console.log(unique);|1842278|1842279|entityMap^0|0|0|0|0|9|6|0|2|1|4|1|A|8|0|2|1|4|1|0|G|1|K|1|10|6|0|E|6|N|6|0|0|0|0|C|7|K|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1D|8|@]|9|@]|A|$]]|$1|E|3|F|5|D|7|1E|8|@]|9|@]|A|$]]|$1|G|3|H|5|D|7|1F|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1G|8|@$K|1H|L|1I|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|1J|8|@$K|1K|L|1L|M|N]|$K|1M|L|1N|M|N]|$K|1O|L|1P|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1Q|8|@$K|1R|L|1S|M|N]|$K|1T|L|1U|M|N]]|9|@]|A|$]]|$1|S|3|T|5|D|7|1V|8|@$K|1W|L|1X|M|N]|$K|1Y|L|1Z|M|N]|$K|20|L|21|M|N]]|9|@]|A|$]]|$1|U|3|V|5|D|7|22|8|@$K|23|L|24|M|N]|$K|25|L|26|M|N]]|9|@]|A|$]]|$1|W|3|X|5|6|7|27|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|10|7|28|8|@]|9|@]|A|$11|12]]|$1|13|3|X|5|6|7|29|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|2A|8|@$K|2B|L|2C|M|N]|$K|2D|L|2E|M|N]]|9|@]|A|$]]|$1|16|3|X|5|6|7|2F|8|@]|9|@]|A|$]]|$1|17|3|18|5|10|7|2G|8|@]|9|@]|A|$11|12]]|$1|19|3|X|5|6|7|2H|8|@]|9|@]|A|$]]|$1|1A|3|-4|5|6|7|2I|8|@]|9|@]|A|$]]]|1B|$]]

<p>Some changes:</p>

<ul>
<li>Move all variable declarations inside of the function.</li>
<li>Use a function parameter for the handed over array, keep the function pure.</li>
<li>Declare all needed variables at top of the function in advance.</li>
<li>Take an array as result array <code>unique</code>.</li>
<li>Check <code>i</code> and <code>j</code> and if equal <code>continue</code> the (inner) loop.</li>
<li>Check the value at <code>i</code> and <code>j</code> and exit the (inner) loop, because a duplicate is found.</li>
<li>Take the check at the end of the inner loop and check the index <code>j</code> with the length of the array <code>l</code>, and if equal push the value to <code>unique</code>.</li>
<li>Use a single <code>return</code> statement with <code>unique</code> array at the end of the outer loop.</li>
</ul>

<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 getUnique(array) {
    var l = array.length,
        i, j,
        unique = [];

    for (i = 0; i &lt; l; i++) {
        for (j = 0; j &lt; l; j++) {
            if (i === j) {
                continue;
            }
            if (array[i] === array[j]) {
                break;
            }
        }
        if (j === l) {
            unique.push(array[i]);
        }
    }
    return unique;
}

console.log(getUnique([-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3]));</code></pre>
</div>
</div>
</p>

<p>Another solution could be to check if <code>indexOf</code> and <code>lastIndexOf</code> returns the same value. Then you found a unique value.</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>var array = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3],
    unique = array.filter((v, i) =&gt; array.indexOf(v) === array.lastIndexOf(v));

console.log(unique);</code></pre>
</div>
</div>
</p>


blocks|key|1967638|text|var+arr1+=+[45,+4,16,25,45,4,16,+9,7,+16,+25];
+var+arr=arr1.sort();
++console.log(arr);
+var+str=[];
+arr.filter(function(value){

if(+arr.indexOf(value)+===+arr.lastIndexOf(value))
{+str.push(value);
console.log("ntttttttttttttnnn"+%2Bstr)

}//+how+this+works+===============A
+})|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1967639|O/P|unstyled|1967640|+7,9|1967641|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>var arr1 = [45, 4,16,25,45,4,16, 9,7, 16, 25];
 var arr=arr1.sort();
  console.log(arr);
 var str=[];
 arr.filter(function(value){

if( arr.indexOf(value) === arr.lastIndexOf(value))
{ str.push(value);
console.log("ntttttttttttttnnn" +str)

}// how this works ===============A
 })
</code></pre>

<p>O/P</p>

<pre><code> 7,9
</code></pre>


blocks|key|1967645|text|Please+try+the+below+code+snippet.

var+arr+=+[-1,+2,+5,+6,+2,+9,+-1,+6,+5,+-1,+3];
++++var+uniqArr+=+[];
++++for+(var+i+=+0;+i+<+arr.length;+i%2B%2B)+{
++++++for+(var+j+=+0;+j+<+arr.length;+j%2B%2B)+{
++++++++if+(arr[i]+==+arr[j]+&&+i+!=+j)+break;
++++++++else+if+(j+==+arr.length+-+1){
++++++++++uniqArr.push(arr[i])
++++++++}+
++++++}
++++}

++++console.log(uniqArr)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1967646|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>Please try the below code snippet.

var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
    var uniqArr = [];
    for (var i = 0; i &lt; arr.length; i++) {
      for (var j = 0; j &lt; arr.length; j++) {
        if (arr[i] == arr[j] &amp;&amp; i != j) break;
        else if (j == arr.length - 1){
          uniqArr.push(arr[i])
        } 
      }
    }

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


blocks|key|1967672|text|这个ES6代码适用于我：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1967673|a.map(c=>a.filter(b=>c==b)).filter(e=>e.length<2).reduce((total，cur)=>+total.concat(cur)，[]|1967674|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

<p>this ES6 code worked for me :</p>

<p>a.map(c=>a.filter(b=>c==b)).filter(e=>e.length&lt;2).reduce((total, cur)=> total.concat(cur), [])</p>


blocks|key|732735|text|对于这个解决方案，可能有一种更简洁的方法，但是通过过滤数组并将其当前值与数组项本身(expect+current+item的索引值)进行比较，可以实现预期的效果。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732736|​|732737|const+sampleArray+=+[1,2,3,7,2,1,3];
const+getNonDuplicatedValues+=+(arr)+=>+
++++arr.filter((item,index)+=>+{
++++++arr.splice(index,1)
++++++const+unique+=+!arr.includes(item)
++++++arr.splice(index,0,item)
++++++return+unique
++})

console.log("Non+duplicated+values:+"+,+...getNonDuplicatedValues(sampleArray))|code-block|syntax|javascript|732738|732739|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>There is possibly a more neat approach to this solution, but this works as expected by filtering the array and compare it's current value with the array items itself (expect current item's index value).</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 sampleArray = [1,2,3,7,2,1,3];
const getNonDuplicatedValues = (arr) =&gt; 
    arr.filter((item,index) =&gt; {
      arr.splice(index,1)
      const unique = !arr.includes(item)
      arr.splice(index,0,item)
      return unique
  })

console.log("Non duplicated values: " , ...getNonDuplicatedValues(sampleArray))</code></pre>
</div>
</div>
</p>


blocks|key|1842371|text|下面是一种循环的工作方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1842372|​|1842373|++var+arr+=+[-1,2,5,6,2,9,-1,6,5,-1,3];
++var+len+=+arr.length;
++const+result+=+arr
++++++++.filter(value=>{
++++++++++++++var+count=0;
++++++++++++++for(var+i=0;i<len;i%2B%2B)
++++++++++++++{
++++++++++++++++++if(arr[i]===value)
++++++++++++++++++++count%2B%2B;
++++++++++++++}
++++++++++++++return+count===1;
++++++++++})
++console.log(result);|code-block|syntax|javascript|1842374|1842375|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>Here is a working method with loops.</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>  var arr = [-1,2,5,6,2,9,-1,6,5,-1,3];
  var len = arr.length;
  const result = arr
        .filter(value=&gt;{
              var count=0;
              for(var i=0;i&lt;len;i++)
              {
                  if(arr[i]===value)
                    count++;
              }
              return count===1;
          })
  console.log(result);</code></pre>
</div>
</div>
</p>


blocks|key|732771|text|另一个简单的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732772|​|732773|var+arr+=+[1,1,2,3,3,4,4,5];

let+duplicateArr+=+[];
var+repeatorCheck+=+(item)+=>+{
++const+currentItemCount+=+arr.filter(val+=>+val===+item).length;
++if(currentItemCount+>+1)+duplicateArr.push(item);
++return+currentItemCount;
}
var+result+=+arr.filter((item,index)+=>+{
++var+itemRepeaterCheck+=+!duplicateArr.includes(item)+&&+repeatorCheck(item);
++if(itemRepeaterCheck+===+1){
++++return+item;
++}
});
console.log(result);|code-block|syntax|javascript|732774|732775|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>Another simple 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>var arr = [1,1,2,3,3,4,4,5];

let duplicateArr = [];
var repeatorCheck = (item) =&gt; {
  const currentItemCount = arr.filter(val =&gt; val=== item).length;
  if(currentItemCount &gt; 1) duplicateArr.push(item);
  return currentItemCount;
}
var result = arr.filter((item,index) =&gt; {
  var itemRepeaterCheck = !duplicateArr.includes(item) &amp;&amp; repeatorCheck(item);
  if(itemRepeaterCheck === 1){
    return item;
  }
});
console.log(result);</code></pre>
</div>
</div>
</p>


blocks|key|1842425|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1842426|const+sampleArr+=+[-1,+2,+5,+6,+2,+9,+-1,+6,+5,+-1,+3];

function+getUnique(arr){
++const+result=[]
++const+obj={}
++for(let+i=0;i<arr.length;i%2B%2B){
++++if(!obj[arr[i]]){
++++++obj[arr[i]]=true
++++++result.push(arr[i])
++++}else{
++++++const+index=+result.indexOf(arr[i])
++++++if(index!==-1){
++++++++result.splice(result.indexOf(arr[i]),1)
++++++}
++++}
++}
++return+result
}

const+uniqueArr=+getUnique(sampleArr)
console.log(uniqueArr)|code-block|syntax|javascript|1842427|1842428|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|K|8|@]|9|@]|A|$E|F]]|$1|G|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|M|8|@]|9|@]|A|$]]]|I|$]]

<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 sampleArr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];

function getUnique(arr){
  const result=[]
  const obj={}
  for(let i=0;i&lt;arr.length;i++){
    if(!obj[arr[i]]){
      obj[arr[i]]=true
      result.push(arr[i])
    }else{
      const index= result.indexOf(arr[i])
      if(index!==-1){
        result.splice(result.indexOf(arr[i]),1)
      }
    }
  }
  return result
}

const uniqueArr= getUnique(sampleArr)
console.log(uniqueArr)</code></pre>
</div>
</div>
</p>


blocks|key|732844|text|这是解决办法..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732845|​|732846|var+x+=+[1,1,2,3,2,4]
var+res+=+[]
x.map(d+=>+{
++if(res.includes(d))+{
++++//+remove+value+from+array
++++res+=+res.filter((a)+=>+a!=d)
++}+else+{
++++//+add+value+to+array
++++res.push(d)
++}
})
console.log(res)+//+[3,4]|code-block|syntax|javascript|732847|732848|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>Here is the solution..</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>var x = [1,1,2,3,2,4]
var res = []
x.map(d =&gt; {
  if(res.includes(d)) {
    // remove value from array
    res = res.filter((a) =&gt; a!=d)
  } else {
    // add value to array
    res.push(d)
  }
})
console.log(res) // [3,4]</code></pre>
</div>
</div>
</p>


blocks|key|1842449|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1842450|let+arr+=+[1,+2,+1,+3,+3,+5];

++++function+nonRepeatableNo(arr)+{
++++++++let+val+=+[]
++++++++for+(let+i+=+0;+i+<+arr.length;+i%2B%2B)+{
++++++++++++let+count+=+0;
++++++++++++for+(let+j+=+0;+j+<+arr.length;+j%2B%2B)+{
++++++++++++++++if+(arr[i]+===+arr[j])+{
++++++++++++++++++++count+%2B=+1
++++++++++++++++}
++++++++++++}
++++++++++++if+(count+===+1)+{
++++++++++++++++val.push(arr[i])
++++++++++++}
++++++++}
++++++++console.log(val)
++++}
++++nonRepeatableNo(arr)|code-block|syntax|javascript|1842451|1842452|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|K|8|@]|9|@]|A|$E|F]]|$1|G|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|M|8|@]|9|@]|A|$]]]|I|$]]

<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>let arr = [1, 2, 1, 3, 3, 5];

    function nonRepeatableNo(arr) {
        let val = []
        for (let i = 0; i &lt; arr.length; i++) {
            let count = 0;
            for (let j = 0; j &lt; arr.length; j++) {
                if (arr[i] === arr[j]) {
                    count += 1
                }
            }
            if (count === 1) {
                val.push(arr[i])
            }
        }
        console.log(val)
    }
    nonRepeatableNo(arr)</code></pre>
</div>
</div>
</p>


<p>Dear all I'm trying to find non repeated value in an array using javascript.I have written some code but it not working properly ..can you guys tell me where is the problem.thanks.     </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>var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
var n = arr.length;
var result = '';

function nonrep() {
  for (var i = 0; i &lt; n; i++) {
    var j;
    for (j = 0; j &lt; n; j++)
      if (i != j &amp;&amp; arr[i] == arr[j]) {
        result = arr[i];
        break;
      }
    if (j == n)
      return arr[i];
  }
  return result;
}
console.log(nonrep())</code></pre>
</div>
</div>
</p>


How to find non repeated numbers in an Array using JavaScript?

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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亲爱的，我试着用javascript在数组中找到非重复值，我已经编写了一些代码，但是它不能正常工作，..can，你们告诉我problem.thanks在哪里。​var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];var n = arr.length;var result = '...

问如何使用JavaScript在数组中查找非重复数字？
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问如何使用JavaScript在数组中查找非重复数字？EN