entityMap|blocks|key|nod5|text|最坏的情况是复杂度是n+*+log(n)。用于循环series+array中所有元素的n和用于二进制搜索的log(n)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|44tk6|​|bss11|const+positions+=+[0,+3,+4,+7,+8];
const+series+=+[0,+4,+8];

function+bsearch(Arr,+value)+{
++var+low+=+0,
++++high+=+Arr.length+-+1,
++++mid;
++while+(low+<=+high)+{
++++mid+=+Math.floor((low+%2B+high)+/+2);
++++if+(Arr[mid]+==+value)+return+mid;
++++else+if+(Arr[mid]+<+value)+low+=+mid+%2B+1;
++++else+high+=+mid+-+1;
++}
++return+-1;
}
let+i;
for+(i+=+0;+i+<+series.length;+%2B%2Bi)+{
++const+pos+=+bsearch(positions,+series[i]);
++if+(pos+===+-1)+break;
}

if+(i+===+series.length)+{
++console.log("All+element+present");
}+else+{
++console.log("All+elmenet+not+present");
}|code-block|syntax|javascript|6gadv|f5v9u^0|A|A|P|C|17|1|1H|6|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|P|9|@$A|Q|B|R|C|D]|$A|S|B|T|C|D]|$A|U|B|V|C|D]|$A|W|B|X|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|Y|9|@]|E|@]|F|$]]|$2|I|4|J|6|K|8|Z|9|@]|E|@]|F|$L|M]]|$2|N|4|H|6|7|8|10|9|@]|E|@]|F|$]]|$2|O|4|-4|6|7|8|11|9|@]|E|@]|F|$]]]]

<p>worst case complexity will be <code>n * log(n)</code>. <code>n</code> for looping all elements in <code>series array</code> and <code>log(n)</code> for binary search.</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 positions = [0, 3, 4, 7, 8];
const series = [0, 4, 8];

function bsearch(Arr, value) {
  var low = 0,
    high = Arr.length - 1,
    mid;
  while (low &lt;= high) {
    mid = Math.floor((low + high) / 2);
    if (Arr[mid] == value) return mid;
    else if (Arr[mid] &lt; value) low = mid + 1;
    else high = mid - 1;
  }
  return -1;
}
let i;
for (i = 0; i &lt; series.length; ++i) {
  const pos = bsearch(positions, series[i]);
  if (pos === -1) break;
}

if (i === series.length) {
  console.log("All element present");
} else {
  console.log("All elmenet not present");
}</code></pre>
</div>
</div>
</p>


entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://github.com/tc39/proposal-set-methods|blocks|key|7f07o|text|Javascript+Set+object缺乏集合论方法，从好的方面来说，它们很容易实现：|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|8arln|​|3872p|class+RealSet+extends+Set+{
++++isSuperSet(iterable)+{
++++++++for+(let+x+of+iterable)+{
++++++++++++if+(!this.has(x))
++++++++++++++++return+false;
++++++++}
++++++++return+true;
++++}
++++//+@TODO:+union,+intersect,+isSubSet,+etc.
}

series+=+[0,4,8]
positions+=+[0,3,4,7,8]

console.log(new+RealSet(positions).isSuperSet(series))|code-block|syntax|javascript|45jit|42cvk|另请参阅:+set方法的TC39+proposal。|dimdf^0|B|3|0|0|0|0|C|D|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|X|G|@$H|Y|I|Z|J|K]]|L|@]|6|$]]|$A|M|C|N|2|E|F|10|G|@]|L|@]|6|$]]|$A|O|C|P|2|Q|F|11|G|@]|L|@]|6|$R|S]]|$A|T|C|N|2|E|F|12|G|@]|L|@]|6|$]]|$A|U|C|V|2|E|F|13|G|@]|L|@$H|14|I|15|A|16]]|6|$]]|$A|W|C|-4|2|E|F|17|G|@]|L|@]|6|$]]]]

<p>Javascript <code>Set</code> object lacks set-theoretical methods, on the bright side, they are easy to implement:</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>class RealSet extends Set {
    isSuperSet(iterable) {
        for (let x of iterable) {
            if (!this.has(x))
                return false;
        }
        return true;
    }
    // @TODO: union, intersect, isSubSet, etc.
}

series = [0,4,8]
positions = [0,3,4,7,8]

console.log(new RealSet(positions).isSuperSet(series))</code></pre>
</div>
</div>
</p>
<p>See also: <a href="https://github.com/tc39/proposal-set-methods" rel="nofollow noreferrer">TC39 proposal</a> for set methods.</p>


entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/every|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf|2|3|blocks|key|414gt|text|我的答案是使用Javascript内置的Array.prototype.every和Array.prototype.indexOf。|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|4kgb2|它会是这样的，|1aqo6|//+Returns+true+if+all+elements+in+firstArray+are+also+in+secondArray
firstArray.every(element+=>+secondArray.indexOf(element)+!==+-1)|code-block|syntax|javascript|fuckb|every+builtin遍历数组，在每个元素上运行回调，如果所有元素的回调返回true+(或truthy值)，则返回true，否则返回false。|863ke|indexOf+builtin在引用相等的数组中搜索您为其提供的参数，如果该参数在数组中，则返回该参数的索引，如果不在，则返回-1。|ac1ln|所以对于firstArray中的每个元素，我们检查它是否在secondArray中，如果是，则通过检查indexOf是否返回-1来返回true。|82qll|结合这些，我们可以在一行程序中解决这个问题。对于未排序的数组，这是您能得到的最好的结果。但是，如果您假设数组已排序，则使用二进制搜索可以更快。正如我面前的答案所使用的。|de2cf^0|K|L|16|N|K|L|0|16|N|1|0|0|0|0|5|0|D|2|0|0|7|0|F|3|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]|9|$2|3|4|5|6|$7|A]]|B|$2|3|4|5|6|$7|8]]|C|$2|3|4|5|6|$7|A]]]|D|@$E|F|G|H|2|I|J|16|K|@$L|17|M|18|N|O]|$L|19|M|1A|N|O]]|P|@$L|1B|M|1C|E|1D]|$L|1E|M|1F|E|1G]]|6|$]]|$E|Q|G|R|2|I|J|1H|K|@]|P|@]|6|$]]|$E|S|G|T|2|U|J|1I|K|@]|P|@]|6|$V|W]]|$E|X|G|Y|2|I|J|1J|K|@$L|1K|M|1L|N|O]]|P|@$L|1M|M|1N|E|1O]]|6|$]]|$E|Z|G|10|2|I|J|1P|K|@$L|1Q|M|1R|N|O]]|P|@$L|1S|M|1T|E|1U]]|6|$]]|$E|11|G|12|2|I|J|1V|K|@]|P|@]|6|$]]|$E|13|G|14|2|I|J|1W|K|@]|P|@]|6|$]]|$E|15|G|-4|2|I|J|1X|K|@]|P|@]|6|$]]]]

<p>My answer to this would be to use Javascript builtins <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/every" rel="nofollow noreferrer"><code>Array.prototype.every</code></a> and <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf" rel="nofollow noreferrer"><code>Array.prototype.indexOf</code></a>.</p>
<p>It would go something like this,</p>
<pre class="lang-js prettyprint-override"><code>// Returns true if all elements in firstArray are also in secondArray
firstArray.every(element =&gt; secondArray.indexOf(element) !== -1)
</code></pre>
<p>The <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/every" rel="nofollow noreferrer"><code>every</code> builtin</a> loops through the array, running your callback on each element and returns true if your callback returned true (or truthy values) for all elements, otherwise returns false.</p>
<p>The <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf" rel="nofollow noreferrer"><code>indexOf</code> builtin</a> searches for the argument you give it in the array with reference equality, returns the index of the argument if it is in the array, or returns -1 if it isn't.</p>
<p>So for every element in firstArray we check if it is in secondArray and then return true if it is by checking if indexOf returns -1 or not.</p>
<p>Combining these, we can solve this problem in a one-liner. For unsorted arrays, this is the best you can get. However, if you assume the array is sorted, you can get much faster using binary search. As the answer before me uses.</p>


entityMap|blocks|key|abp1k|text|如果您的数组(这两个数组)是排序的，并且它们不包含重复项，则可以通过以下方式遍历它们：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6ve0j|​|f4in|//const+series+=+[0,4,8];
const+positions+=+[0,3,4,7,8];

function+check(series,positions){
++let+pos=0;
++for(let+item+of+series){
++++while(positions[pos]<item+&&+pos<positions.length)
++++++pos%2B%2B;
++++if(pos==positions.length+%7C%7C+positions[pos]!==item)
++++++return+false;
++}
++return+true;
}

console.log("true?",check([0,4,8],positions));
console.log("true?",check([0],positions));
console.log("false?",check([5],positions));
console.log("true?",check([8],positions));
console.log("false?",check([0,4,5],positions));|code-block|syntax|javascript|7t1ko|e96ol^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|M|9|@]|A|@]|B|$]]|$2|E|4|F|6|G|8|N|9|@]|A|@]|B|$H|I]]|$2|J|4|D|6|7|8|O|9|@]|A|@]|B|$]]|$2|K|4|-4|6|7|8|P|9|@]|A|@]|B|$]]]]

<p>If your arrays (both of them) are sorted and they don't contain repeats, you can walk through them this way:</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 series = [0,4,8];
const positions = [0,3,4,7,8];

function check(series,positions){
  let pos=0;
  for(let item of series){
    while(positions[pos]&lt;item &amp;&amp; pos&lt;positions.length)
      pos++;
    if(pos==positions.length || positions[pos]!==item)
      return false;
  }
  return true;
}

console.log("true?",check([0,4,8],positions));
console.log("true?",check([0],positions));
console.log("false?",check([5],positions));
console.log("true?",check([8],positions));
console.log("false?",check([0,4,5],positions));</code></pre>
</div>
</div>
</p>


<p>Let's say I have an array <code>const series = [0,4,8]</code> and another array <code>const positions = [0,3,4,7,8]</code></p>
<p>How would I detect if all of the values of <code>series</code> are included in <code>positions</code>?</p>
<p>My initial reaction was to make a <code>for loop</code> that would cycle through all the values in <code>positions</code> and check to see if they equal the values of <code>series</code> but that seems highly inefficient.</p>


Javascript Detect All Values of Array in another Array

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。

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 假设我有一个数组const series = [0,4,8]和另一个数组const positions = [0,3,4,7,8] 如何检测positions中是否包含series的所有值 我最初的反应是创建一个for loop，它将遍历positions中的所有值，并检查它们是否等于series的值，但这似乎非常低...

问Javascript检测另一个数组中数组的所有值
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