blocks|key|1407729|text|使用以下内容更改代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1407730|++++var+numbers4+=+[5,+2,+3,+4,+2,+6,+7,+1,+2,+3];
++++var+firstIndex+=+"";
+++var+isMatch=false;
++++for+(var+a+=+0;+a+<+numbers4.length;+a%2B%2B)+{
++++++++for+(var+b+=+a%2B1;+b+<+numbers4.length;+b%2B%2B)+{
++++++++++++if+(numbers4[a]+===+numbers4[b]){
++++++++++++++++firstIndex+=+numbers4.indexOf(numbers4[a]);
++++++++++++++++isMatch=true;
++++++++++++++++break;
++++++++++}
++++++++}
+++++++++++if+(isMatch)+{break;}
++++}
++++console.log(firstIndex);|code-block|syntax|javascript|1407731|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>Change your code with the following</p>

<pre><code>    var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
    var firstIndex = "";
   var isMatch=false;
    for (var a = 0; a &lt; numbers4.length; a++) {
        for (var b = a+1; b &lt; numbers4.length; b++) {
            if (numbers4[a] === numbers4[b]){
                firstIndex = numbers4.indexOf(numbers4[a]);
                isMatch=true;
                break;
          }
        }
           if (isMatch) {break;}
    }
    console.log(firstIndex);
</code></pre>


blocks|key|2284536|text|我会用一个对象来记住已经找到的值..。这样的事情应该有效;)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2284537|​|2284538|var+numbers4+=+[5,+2,+3,+4,+4,+6,+7,+1,+2,+3];

function+findFirstDuplicateIndex(arr){
++var+found+=+{};
++
++for+(var+a+=+0,+aa+=+arr.length;+a+<+aa+;+a%2B%2B)+{
++++if+(found[arr[a]])
++++++return+found[arr[a]];
++
++++found[numbers4[a]]+=+a
++}
}

console.log(findFirstDuplicateIndex(numbers4));|code-block|syntax|javascript|2284539|2284540|它非常快，因为您只循环了一次数组。其余的时间，您只访问对象属性或设置对象属性.(如果你有问题，请告诉我;)|2284541|但是也许有更快的..。这只是一个想法%5E%5E|2284542|PS:它也适用于单词，而不仅仅是数字。|2284543|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|V|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|W|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|X|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|Q|$]]

<p>I would use an object remembering the values already found... Something like that should work ;)</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 numbers4 = [5, 2, 3, 4, 4, 6, 7, 1, 2, 3];

function findFirstDuplicateIndex(arr){
  var found = {};
  
  for (var a = 0, aa = arr.length; a &lt; aa ; a++) {
    if (found[arr[a]])
      return found[arr[a]];
  
    found[numbers4[a]] = a
  }
}

console.log(findFirstDuplicateIndex(numbers4));</code></pre>
</div>
</div>
</p>

<p>It's quite fast because you just loop one time through the array. The rest of the time you just access an object property or you set the object property... Let me know if you have questions ;)</p>

<p>However maybe there something faster... It's just an idea ^^</p>

<p>PS: It also works with words, not just numbers</p>


blocks|key|2692915|text|甚至不需要嵌套for循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2692916|​|2692917|var+numbers4+=+[5,+2,+3,+4,+2,+6,+7,+1,+2,+3];
var+firstIndex+=+"";
for+(var+a+=+1;+a+<+numbers4.length;+a%2B%2B)+{+//start+from+second+elem+since+first+is+never+a+duplicate
++++++++if+(numbers4.lastIndexOf(numbers4[a])>+a)+{
++++++++++++firstIndex+=+a;
++++++++++++break;
++++++++}
}
console.log(firstIndex);+//1|code-block|syntax|javascript|2692918|2692919|在迭代过程中，您所要做的就是检查当前值是否存在于数组中的某个地方。这是通过使用lastIndexOf()检查该值出现的最后一个索引来完成的。|offset|length|style|CODE|2692920|entityMap^0|0|0|0|0|13|D|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|V|8|@$L|W|M|X|N|O]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|Q|$]]

<p>You don't even need nested for 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 numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 1; a &lt; numbers4.length; a++) { //start from second elem since first is never a duplicate
        if (numbers4.lastIndexOf(numbers4[a])&gt; a) {
            firstIndex = a;
            break;
        }
}
console.log(firstIndex); //1</code></pre>
</div>
</div>
</p>

<p>All you have to do during iterating is to check if current value exists somewhere further in array. That is done by checking last index of this value's occurrence using <code>lastIndexOf()</code>. </p>


blocks|key|2284603|text|你的问题是-你有两个循环，只有一个break，你需要打破这两个。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2284604|为什么不在值匹配时立即返回索引呢？|2284605|var+numbers4+=+[5,+2,+3,+4,+2,+6,+7,+1,+2,+3];
function+getDuplicate(numbers4)
{
++++for+(var+a+=+0;+a+<+numbers4.length;+a%2B%2B)+{
++++++++for+(var+b+=+a%2B1;+b+<+numbers4.length;+b%2B%2B)+{
++++++++++++if+(numbers4[a]+===+numbers4[b]){
++++++++++++++++return+a;
++++++++++++}
++++++++}
++++}
}
console.log(getDuplicate(numbers4+));|code-block|syntax|javascript|2284606|但是，您可以通过保持映射来进一步优化代码。|2284607|function+getDuplicate(+numbers+)
{
++++var+map+=+{};
++++for+(var+a+=+0;+a+<+numbers.length;+a%2B%2B)+
++++{
++++++++if(+map[+numbers[+a+]+]+)
++++++++{
++++++++++return+a;
++++++++}
++++++++map[+numbers[+a+]+]+=+true;
++++}
++++return+-1;
}|2284608|entityMap^0|H|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

<p>Your problem is - you have two loops and only one <code>break</code>, you need to break out of both.</p>

<p>Why not simply return the index as soon as values match?</p>

<pre><code>var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
function getDuplicate(numbers4)
{
    for (var a = 0; a &lt; numbers4.length; a++) {
        for (var b = a+1; b &lt; numbers4.length; b++) {
            if (numbers4[a] === numbers4[b]){
                return a;
            }
        }
    }
}
console.log(getDuplicate(numbers4 ));
</code></pre>

<p>However, you can optimize your code further by keeping a map</p>

<pre><code>function getDuplicate( numbers )
{
    var map = {};
    for (var a = 0; a &lt; numbers.length; a++) 
    {
        if( map[ numbers[ a ] ] )
        {
          return a;
        }
        map[ numbers[ a ] ] = true;
    }
    return -1;
}
</code></pre>


blocks|key|1403929|text|您可以检查indexOf()是否等于lastIndexOf()，并返回值和中断循环。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1403930|​|1403931|var+numbers4+=+[5,+2,+3,+4,+2,+6,+7,+1,+2,+3];
var+firstIndex+=+"";
for+(var+i+=+0;+i+<+numbers4.length;+i%2B%2B)+{
++if+(numbers4.indexOf(numbers4[i])+!=+numbers4.lastIndexOf(numbers4[i]))+{
++++firstIndex+=+i;
++++break;
++}
}

console.log(firstIndex)|code-block|syntax|javascript|1403932|1403933|entityMap^0|5|9|I|D|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|V|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|W|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|O|$]]

<p>You can check if <code>indexOf()</code> is not equal to <code>lastIndexOf()</code> and return value and break 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 numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var i = 0; i &lt; numbers4.length; i++) {
  if (numbers4.indexOf(numbers4[i]) != numbers4.lastIndexOf(numbers4[i])) {
    firstIndex = i;
    break;
  }
}

console.log(firstIndex)</code></pre>
</div>
</div>
</p>


blocks|key|1407833|text|您的中断；终止b循环，因为if()没有使用括号{}。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1407834|此外，firstIndex应该简单地设置为a或b，这取决于您需要返回重复的第一次出现还是第一次重复。|1407835|它应该是：|1407836|if+(numbers4[a]+===+numbers4[b])
{
++++++++++++firstIndex+=+a;
++++++++++++break;
}|code-block|syntax|javascript|1407837|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<p>Your break; terminates b loop, because if() is not using parenthesis {}.</p>

<p>Additionally, firstIndex should be simply set to either a or b depending on whether you need to return the duplicate's first occurance or first repetition.</p>

<p>It should be:</p>

<pre><code>if (numbers4[a] === numbers4[b])
{
            firstIndex = a;
            break;
}
</code></pre>


blocks|key|1407863|text|如果我正确理解了你的问题，那会对你有帮助.基本上，您需要进行双重迭代。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1407864|​|1407865|const+firstDupeIndex+=+list+=>+list.findIndex(
++(item,+index)+=>+list.lastIndexOf(item)+!==+index
);

console.log(
++"First+Dupe+at+index:",+
++firstDupeIndex([5,+2,+3,+4,+4,+6,+7,+1,+2,+3])
);|code-block|syntax|javascript|1407866|1407867|上面的实现具有O(n2)的缺点，因为在findIndex函数中嵌套了findIndex。|offset|length|style|CODE|1407868|一个更好的解决方案是通过构建一个字典来索引所出现的情况，从而在最坏的情况下将时间复杂度保持在O(n)上。可能有点不那么整洁，但在大量投入的情况下肯定会更有效率。|1407869|1407870|const+firstDupeIndex+=+(list)+=>+{
++const+dict+=+{};
++
++for+(const+[index,+value]+of+list.entries())+{
++++if+(dict.hasOwnProperty(value))+{
++++++return+dict[value];
++++}
++
++++dict[value]+=+index;
++}
++
++return+-1;
};

console.log(
++"First+Dupe+at+index:",+
++firstDupeIndex(['a',+'b',+'c',+'d',+'e',+'b',+'z',+'t',+'c'])
);|1407871|1407872|entityMap^0|0|0|0|0|7|5|J|9|Y|9|0|1A|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|10|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|11|8|@$L|12|M|13|N|O]|$L|14|M|15|N|O]|$L|16|M|17|N|O]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|18|8|@$L|19|M|1A|N|O]]|9|@]|A|$]]|$1|R|3|C|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|1C|8|@]|9|@]|A|$G|H]]|$1|U|3|C|5|6|7|1D|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|W|$]]

<p>If I correctly understand your question, that's should help you...
Basically, you need for a double iteration.</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 firstDupeIndex = list =&gt; list.findIndex(
  (item, index) =&gt; list.lastIndexOf(item) !== index
);

console.log(
  "First Dupe at index:", 
  firstDupeIndex([5, 2, 3, 4, 4, 6, 7, 1, 2, 3])
);</code></pre>
</div>
</div>
</p>
<hr />
<p>Thee above implementation comes with the drawback of being <code>O(n2)</code>, due to nesting the <code>lastIndexOf</code> within the <code>findIndex</code> function.</p>
<p>A better solution would be to index your occurrences by building a dictionary, therefore keeping time complexity to just <code>O(n)</code> in the worst case. Probably a little bit less neat, but surely more efficient with big inputs.</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 firstDupeIndex = (list) =&gt; {
  const dict = {};
  
  for (const [index, value] of list.entries()) {
    if (dict.hasOwnProperty(value)) {
      return dict[value];
    }
  
    dict[value] = index;
  }
  
  return -1;
};

console.log(
  "First Dupe at index:", 
  firstDupeIndex(['a', 'b', 'c', 'd', 'e', 'b', 'z', 't', 'c'])
);</code></pre>
</div>
</div>
</p>


blocks|key|2693167|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2693168|var+numbers+=+[5,+2,+3,+4,+2,+6,+7,+1,+2,+3];
var+firstIndex;
var+len+=+numbers.length;

for+(var+i+=+0;+i+<+len;+i%2B%2B)+{
++var+tmpArray+=+numbers.slice(i+%2B+1,+len);
++var+index+=+tmpArray.indexOf(numbers[i]);
++if+(index+!==+-1)+{
++++firstIndex+=+index+%2B+i+%2B+1;
++++break;
++}
}
console.log(firstIndex);|code-block|syntax|javascript|2693169|2693170|更新：|offset|length|style|BOLD|2693171|实际上，您的逻辑是正确的，但是您忽略了用于if+condition的大括号，而且如果条件满足，则意味着firstIndex与a相同。|CODE|2693172|这是你带大括号的代码，|2693173|2693174|var+numbers4+=+[5,+2,+3,+4,+2,+6,+7,+1,+2,+3];
var+firstIndex+=+"";
for+(var+a+=+0;+a+<+numbers4.length;+a%2B%2B)+{
++for+(var+b+=+a+%2B+1;+b+<+numbers4.length;+b%2B%2B)+{
++++if+(numbers4[a]+===+numbers4[b])+{
++++++firstIndex+=+a
++++++break;
++++}
++}
}
console.log(firstIndex);|2693175|2693176|entityMap^0|0|0|0|0|3|0|L|C|1F|A|1Q|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|11|8|@$J|12|K|13|L|M]]|9|@]|A|$]]|$1|N|3|O|5|6|7|14|8|@$J|15|K|16|L|P]|$J|17|K|18|L|P]|$J|19|K|1A|L|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|A|$]]|$1|S|3|4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|T|3|U|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|V|3|4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|X|$]]

<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 numbers = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex;
var len = numbers.length;

for (var i = 0; i &lt; len; i++) {
  var tmpArray = numbers.slice(i + 1, len);
  var index = tmpArray.indexOf(numbers[i]);
  if (index !== -1) {
    firstIndex = index + i + 1;
    break;
  }
}
console.log(firstIndex);</code></pre>
</div>
</div>
</p>

<p><strong>Update:</strong></p>

<p>Actually your logic is right, but you missed braces for <code>if condition</code> and also if the condition satisfies then it means <code>firstIndex</code> is the same as <code>a</code></p>

<p>This is your code with braces,</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 numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 0; a &lt; numbers4.length; a++) {
  for (var b = a + 1; b &lt; numbers4.length; b++) {
    if (numbers4[a] === numbers4[b]) {
      firstIndex = a
      break;
    }
  }
}
console.log(firstIndex);</code></pre>
</div>
</div>
</p>


blocks|key|2284780|text|这个问题指出数组中的第一个dupe必须和它的索引一起找到。因此，我返回一个对象，其中i属性是index，e属性是第一个重复的e功能本身。执行此任务的一种方法是|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|2284781|​|2284782|var+numbers4+=+[5,+2,+3,+4,+2,+6,+7,+1,+2,+3],
++++headDupe+=+(a,r={})+=>+(r.e+=+a.find((n,i)+=>+(r.i+=+a.indexOf(n),+r.i+<+i)),r);

console.log(headDupe(numbers4));|code-block|syntax|javascript|2284783|2284784|entityMap^0|16|1|1G|1|1A|1|1Q|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|D]|$9|X|A|Y|B|D]]|E|@]|F|$]]|$1|G|3|H|5|6|7|Z|8|@]|E|@]|F|$]]|$1|I|3|J|5|K|7|10|8|@]|E|@]|F|$L|M]]|$1|N|3|H|5|6|7|11|8|@]|E|@]|F|$]]|$1|O|3|-4|5|6|7|12|8|@]|E|@]|F|$]]]|P|$]]

<p>The question states the first dupe in the array has to be found along with it's index. So I return an object where the <code>i</code> property is the <strong>i</strong>ndex and the <code>e</code> property is the first duplicate <strong>e</strong>lement itself. One way of performing this task would be</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 numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3],
    headDupe = (a,r={}) =&gt; (r.e = a.find((n,i) =&gt; (r.i = a.indexOf(n), r.i &lt; i)),r);

console.log(headDupe(numbers4));</code></pre>
</div>
</div>
</p>


<p>I have to find first duplicate value in array and then return its index in variable firstIndex. This has to be done with for loop which should stop after having found first duplicate. I know this is probably pretty simple but I got stuck. So far I have this but it doesn't seem to be working:</p>

<pre><code>var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 0; a &lt; numbers4.length; a++) {
    for (var b = a+1; b &lt; numbers4.length; b++) {
        if (numbers4[a] === numbers4[b])
            firstIndex = numbers4.indexOf(numbers4[a]);
            break;
    }
}
console.log(firstIndex);
</code></pre>

<p>Console prints out 1 which is fine because 2 is first duplicate, but when I change numbers in array, the loop doesn't work. Can you advise what can be changed here?</p>

<p>Thanks in advance!</p>


Javascript: How to find first duplicate value and return its index?

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解答不清晰，无法理解或解决问题。

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我必须在数组中找到第一个重复值，然后在变量firstIndex中返回它的索引。这必须用for循环来完成，在找到第一个重复之后应该停止。我知道这可能很简单但我被困住了。到目前为止，我有这个，但它似乎不起作用：var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];var firstIn...

问Javascript:如何找到第一个重复值并返回其索引？
EN

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问Javascript:如何找到第一个重复值并返回其索引？EN