blocks|key|2497079|text|我将使用一个Map，它的迭代方法(这里是entries())按插入顺序迭代。将计数加到Map中，类似于您正在做的操作，然后获取Map的条目，根据计数为1的条目进行过滤，然后进行切片以获得前两个条目：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2497080|​|2497081|var+list+=+[5,+23,+23,+5,+66,+3,+23,+7,+3,+0];
const+countByNum+=+new+Map();
list.forEach((num)+=>+{
++countByNum.set(num,+(countByNum.get(num)+%7C%7C+0)+%2B+1);
});
const+firstTwoNonDupes+=+[...countByNum.entries()]
++.filter(([,+count])+=>+count+===+1)
++.slice(0,+2)
++.map(([num])+=>+num);
console.log(firstTwoNonDupes);|code-block|syntax|javascript|2497082|2497083|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map/entries^0|K|9|K|9|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]]]

<p>I'd use a Map instead, whose iteration methods (here, <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map/entries" rel="nofollow noreferrer"><code>entries()</code></a>) iterate in insertion order. Count up into a Map, similar to what you're doing, then get the Map's entries, filter by those with a count of 1, and slice to get the first two:</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 list = [5, 23, 23, 5, 66, 3, 23, 7, 3, 0];
const countByNum = new Map();
list.forEach((num) =&gt; {
  countByNum.set(num, (countByNum.get(num) || 0) + 1);
});
const firstTwoNonDupes = [...countByNum.entries()]
  .filter(([, count]) =&gt; count === 1)
  .slice(0, 2)
  .map(([num]) =&gt; num);
console.log(firstTwoNonDupes);</code></pre>
</div>
</div>
</p>


blocks|key|2497121|text|您可以首先将数字映射为键，将重复映射为值，然后在数组上再次循环，并在映射对象上选择值为1的前两个数字|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2497122|var+list+=+[5,+23,+23,+5,+66,+3,+23,+7,+3,+0];

let+mapping+=+list.reduce((op,inp)=>{
++op[inp]+=+op[inp]+%7C%7C+0
++op[inp]%2B%2B
++return+op
},{})

let+firstTwo+=+[]

list.forEach(v=>+{
++if(mapping[v]+===+1+&&+firstTwo.length+<+2){
++++firstTwo.push(v)
++}
})

console.log(firstTwo);|code-block|syntax|javascript|2497123|​|2497124|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>You can first map digits as key and repetition as value, then loop again on array and select first two digits which has value as 1 on mapping object
<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 list = [5, 23, 23, 5, 66, 3, 23, 7, 3, 0];

let mapping = list.reduce((op,inp)=&gt;{
  op[inp] = op[inp] || 0
  op[inp]++
  return op
},{})

let firstTwo = []

list.forEach(v=&gt; {
  if(mapping[v] === 1 &amp;&amp; firstTwo.length &lt; 2){
    firstTwo.push(v)
  }
})

console.log(firstTwo);</code></pre>
</div>
</div>
</p>


blocks|key|2284516|text|您可以采用Map和Set的组合，其中映射只存储已见的值，并将其作为可见的值的结果集。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2284517|​|2284518|function+getFirstTwoNonDuplicateValues(array)+{
++++var+map+=+new+Map,
++++++++result+=+new+Set,
++++++++value;
++++
++++for+(value+of+array)+map.set(value,+map.has(value)
++++++++?+result.delete(value)
++++++++:+result.add(value));
++++
++++return+[...result].slice(0,+2);
}

console.log(getFirstTwoNonDuplicateValues([5,+23,+23,+5,+66,+3,+23,+7,+3,+0]));|code-block|syntax|javascript|2284519|2284520|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set^0|5|3|9|3|5|3|0|9|3|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]|$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|1A|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|O|$P|$5|Q|R|S|E|$T|U]]|V|$5|Q|R|S|E|$T|W]]]]

<p>You could take a combination of a <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map" rel="nofollow noreferrer"><code>Map</code></a>, where the map stores just the seen values and a <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set" rel="nofollow noreferrer"><code>Set</code></a> as result set for value who are seen ones.</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 getFirstTwoNonDuplicateValues(array) {
    var map = new Map,
        result = new Set,
        value;
    
    for (value of array) map.set(value, map.has(value)
        ? result.delete(value)
        : result.add(value));
    
    return [...result].slice(0, 2);
}

console.log(getFirstTwoNonDuplicateValues([5, 23, 23, 5, 66, 3, 23, 7, 3, 0]));</code></pre>
</div>
</div>
</p>


<p>I encountered this problem in a javascript test wherein I need to print first 2 non duplicate numbers and here is my partial attempt.</p>

<pre><code>var list = [5, 23, 23, 5, 66, 3, 23, 7, 3, 0];


var locObj = {};
var count = 1;

for(var i =0; i&lt; list.length; i++){
  if(!(list[i] in locObj))
  locObj[list[i]] = 1;
  else
    locObj[list[i]] = count++;
}

console.log(locObj);//has an object with value as 1 if its non duplicate and the number of occurrences in the case of duplicate values.
</code></pre>

<p>I can further loop through locObj and determine whose value is equal to 1 and then print 2 values. But since javascript keys are not ordered, I'm unable to print 66 and 7 which is the expected out. How to solve it? Appreciate any other better approach to solve the problem as well.</p>


print first 2 non duplicate numbers javascript

我在一个javascript测试中遇到了这个问题，我需要打印前两个非重复的数字，下面是我的部分尝试。var list = [5, 23, 23, 5, 66, 3, 23, 7, 3, 0];var locObj = {};var count = 1;for(var i =0; i< list.length; i++)...

问打印前2个非重复数字javascript
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