blocks|key|4391625|text|让我们编写一个函数，以数组的形式返回字符串的所有排列。由于您不需要任何全局变量，因此返回排列是至关重要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4391626|++function+permut(string)+{
++if+(string.length+<+2)+return+string;+//+This+is+our+break+condition

++var+permutations+=+[];+//+This+array+will+hold+our+permutations
++for+(var+i+=+0;+i+<+string.length;+i%2B%2B)+{
++++var+char+=+string[i];

++++//+Cause+we+don't+want+any+duplicates:
++++if+(string.indexOf(char)+!=+i)+//+if+char+was+used+already
++++++continue;+//+skip+it+this+time

++++var+remainingString+=+string.slice(0,+i)+%2B+string.slice(i+%2B+1,+string.length);+//Note:+you+can+concat+Strings+via+'%2B'+in+JS

++++for+(var+subPermutation+of+permut(remainingString))
++++++permutations.push(char+%2B+subPermutation)
++}
++return+permutations;
}|code-block|syntax|javascript|4391627|要打印它们，只需在之后迭代数组：|4391628|+var+myString+=+"xyz";
+permutations+=+permut(myString);
+for+(permutation+of+permutations)
+++print(permutation)+//Use+the+output+method+of+your+choice|4391629|希望我能帮助你回答你的问题。|4391630|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

<p>Let's write a function that returns all permutations of a string as an array. As you don't want any global variables, returning the permutations is crucial.</p>

<pre><code>  function permut(string) {
  if (string.length &lt; 2) return string; // This is our break condition

  var permutations = []; // This array will hold our permutations
  for (var i = 0; i &lt; string.length; i++) {
    var char = string[i];

    // Cause we don't want any duplicates:
    if (string.indexOf(char) != i) // if char was used already
      continue; // skip it this time

    var remainingString = string.slice(0, i) + string.slice(i + 1, string.length); //Note: you can concat Strings via '+' in JS

    for (var subPermutation of permut(remainingString))
      permutations.push(char + subPermutation)
  }
  return permutations;
}
</code></pre>

<p>To print them, just iterate over the array afterwards:</p>

<pre><code> var myString = "xyz";
 permutations = permut(myString);
 for (permutation of permutations)
   print(permutation) //Use the output method of your choice
</code></pre>

<p>Hope I could help you with your question.</p>


blocks|key|4391654|text|排列问题已经研究到死了。Heap's+algorithm是一个众所周知的解决方案。以下是使用生成器的JS版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4391655|​|4391656|function+*permute(a,+n+=+a.length)+{
++if+(n+<=+1)+yield+a.slice();
++else+for+(let+i+=+0;+i+<+n;+i%2B%2B)+{
++++yield+*permute(a,+n+-+1);
++++const+j+=+n+%25+2+?+0+:+i;
++++[a[n-1],+a[j]]+=+[a[j],+a[n-1]];
++}
}

console.log(Array.from(permute("abcabad".split('')))
.map(perm+=>+perm.join(''))
.filter((el,+idx,+self)+=>+(self.indexOf(el)+===+idx)));|code-block|syntax|javascript|4391657|4391658|permute旨在获取并生成数组，而不是字符串，因此我们在调用字符串之前将其拆分为字符，并在打印结果之前将字符粘贴回字符串中。|style|CODE|4391659|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Heap%2527s_algorithm^0|C|G|0|0|0|0|0|0|7|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|E|5|6|7|11|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|12|8|@]|9|@]|C|$I|J]]|$1|K|3|E|5|6|7|13|8|@]|9|@]|C|$]]|$1|L|3|M|5|6|7|14|8|@$A|15|B|16|N|O]]|9|@]|C|$]]|$1|P|3|-4|5|6|7|17|8|@]|9|@]|C|$]]]|Q|$R|$5|S|T|U|C|$V|W]]]]

<p>The problem of permutations has been studied to death. <a href="https://en.wikipedia.org/wiki/Heap%27s_algorithm" rel="noreferrer">Heap's algorithm</a> is one well-known solution. Here is a version in JS, using a generator:</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 *permute(a, n = a.length) {
  if (n &lt;= 1) yield a.slice();
  else for (let i = 0; i &lt; n; i++) {
    yield *permute(a, n - 1);
    const j = n % 2 ? 0 : i;
    [a[n-1], a[j]] = [a[j], a[n-1]];
  }
}

console.log(Array.from(permute("abcabad".split('')))
.map(perm =&gt; perm.join(''))
.filter((el, idx, self) =&gt; (self.indexOf(el) === idx)));</code></pre>
</div>
</div>
</p>

<p><code>permute</code> is designed to take and generate arrays, not strings, so we split the string into characters before calling it, and paste the characters back into strings before printing out the results.</p>


blocks|key|4391709|text|permutation=(str,prefix)=>{
+++if(str.length==0){
++++++console.log(prefix);
+++}
+++else{
++++++for(let+i=0;i<str.length;i%2B%2B){
++++++++++let+rem+=+str.substring(0,i)%2Bstr.substring(i%2B1);
++++++++++permutation(rem,prefix%2Bstr[i]);
+++++++}
++++}
+}
+let+str="ABC";
+permutation(str,"");|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4391710|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>permutation=(str,prefix)=&gt;{
   if(str.length==0){
      console.log(prefix);
   }
   else{
      for(let i=0;i&lt;str.length;i++){
          let rem = str.substring(0,i)+str.substring(i+1);
          permutation(rem,prefix+str[i]);
       }
    }
 }
 let str="ABC";
 permutation(str,"");
</code></pre>


blocks|key|4391942|text|​|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4391943|var+str+=+"abcdefgh";

for(let+i+=+0;+i<str.length;+i%2B%2B){
++++for(let+j+=+i;+j<=str.length;+j%2B%2B){
++++++++if(i+!=+j){
++++++++++++var+out+=+str.slice(i,j);
++++++++console.log(out);
++++++++}
++++++++
++++}
}|code-block|syntax|javascript|4391944|4391945|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>var str = "abcdefgh";

for(let i = 0; i&lt;str.length; i++){
    for(let j = i; j&lt;=str.length; j++){
        if(i != j){
            var out = str.slice(i,j);
        console.log(out);
        }
        
    }
}</code></pre>
</div>
</div>
</p>


blocks|key|4391950|text|const+permut+=+(str)+=>+{
++if+(str.length+<=+2)+return+str.length+===+2+?+[str,+str[1]+%2B+str[0]]+:+[str];
++return+str
++++.split("")
++++.reduce(
++++++(acc,+letter,+i)+=>
++++++++acc.concat(
++++++++++permut(str.slice(0,+i)+%2B+str.slice(i+%2B+1)).map((val)+=>+letter+%2B+val)
++++++++),
++++++[]
++++);
};|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4391951|找到here|unstyled|offset|length|4391952|entityMap|0|LINK|mutability|MUTABLE|url|https://www.w3resource.com/javascript-exercises/fundamental/javascript-fundamental-exercise-136.php^0|0|2|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>const permut = (str) =&gt; {
  if (str.length &lt;= 2) return str.length === 2 ? [str, str[1] + str[0]] : [str];
  return str
    .split(&quot;&quot;)
    .reduce(
      (acc, letter, i) =&gt;
        acc.concat(
          permut(str.slice(0, i) + str.slice(i + 1)).map((val) =&gt; letter + val)
        ),
      []
    );
};
</code></pre>
<p>found <a href="https://www.w3resource.com/javascript-exercises/fundamental/javascript-fundamental-exercise-136.php" rel="nofollow noreferrer">here</a></p>


<p>I've seen versions of this question for other languages, but not for JS.</p>

<p>Is it possible to do this recursively in one function?</p>

<p>I understand that I need to take the first element in the string, and then append it to each solution to the recursion on the remainder of the string.
So logically, I understand how the recursion needs to go. I just don't understand how to append the first char onto each of the recursive solutions</p>

<pre><code>var myString = "xyz";

function printPermut(inputString){
    var outputString;
    if(inputString.length === 0){
        return inputString;
    }
    if(inputString.length === 1){
        return inputString;
    }
    else{
       for(int i = 0; i&lt;inputString.length(); i++){
           //something here like: 
           //outputString = outputString.concat(printPermut(inputString.slice(1))??
           //maybe store each unique permutation to an array or something?
       } 
    }
}
</code></pre>


Recursively print all permutations of a string (Javascript)

我已经在其他语言中看到过这个问题的版本，但没有在JS中。是否可以在一个函数中递归地执行此操作？我知道我需要获取字符串中的第一个元素，然后将其附加到字符串其余部分的递归的每个解决方案中。因此，从逻辑上讲，我理解递归需要如何进行。我只是不明白如何将第一个char附加到每个递归解决方案中var myString = "xyz...

问递归打印字符串的所有排列(Javascript)
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