blocks|key|5088767|text|你可以试试这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5088768|var+mergeArrays+=+function(arr1+,+arr2){
++++var+mergedArray+=+arr1.concat(arr2);

++++var+uniqueArray+=+mergedArray.filter(function(elem,+pos)+{
++++++++return+mergedArray.indexOf(elem)+==+pos;
++++});
++++return+uniqueArray;
}+|code-block|syntax|javascript|5088769|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|$]]

You can try this

<pre><code>var mergeArrays = function(arr1 , arr2){
 var mergedArray = arr1.concat(arr2);

 var uniqueArray = mergedArray.filter(function(elem, pos) {
 return mergedArray.indexOf(elem) == pos;
 });
 return uniqueArray;
} 
</code></pre>

blocks|key|3620707|text|function+merge(...args)+{
++++return+[...new+Set(args.reduce((acc,val)+=>+[...acc,+...val]))].sort();
}+//Short+and+crisp+solution+to+merge+arrays,+sort+them+and+remove+duplicates|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3620708|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>function merge(...args) {
 return [...new Set(args.reduce((acc,val) =&gt; [...acc, ...val]))].sort();
} //Short and crisp solution to merge arrays, sort them and remove duplicates
</code></pre>

blocks|key|5088760|text|如果你对underscore.js是开放的，而且它很棒，那么你可以尝试像_.union这样的东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5088761|http://underscorejs.org/#union|offset|length|5088762|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|U|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@$D|O|E|P|1|Q]]|A|$]]|$1|F|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|C]]]]

If you're open to underscore.js, and it's great, then you could try something like _.union

<a href="http://underscorejs.org/#union" rel="nofollow">http://underscorejs.org/#union</a>

blocks|key|3620703|text|我会这样做的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3620704|function+unique(array)+{
++++var+a+=+array.concat();
++++for(var+i=0;+i<a.length;+%2B%2Bi)+{
++++++++for(var+j=i%2B1;+j<a.length;+%2B%2Bj)+{
++++++++++++if(a[i]+===+a[j])
++++++++++++++++a.splice(j--,+1);
++++++++}
++++}
++++return+a;
};
var+array1+=+["...",",,,"];
var+array2+=+["///",+"..."];
var+array3+=+unique(array1.concat(array2));|code-block|syntax|javascript|3620705|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|$]]

I would do it this way

<pre><code>function unique(array) {
 var a = array.concat();
 for(var i=0; i&lt;a.length; ++i) {
 for(var j=i+1; j&lt;a.length; ++j) {
 if(a[i] === a[j])
 a.splice(j--, 1);
 }
 }
 return a;
};
var array1 = ["...",",,,"];
var array2 = ["///", "..."];
var array3 = unique(array1.concat(array2));
</code></pre>

blocks|key|1248370|text|首先，你的代码并没有做你认为它做的事情。试着用这个例子运行它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1248371|array1+=+[0,0,0,1,1,1,3,5,5,5,5,9];
array2+=+[1,1,1,1,1,1];|code-block|syntax|javascript|1248372|你就会知道哪里出了问题。|1248373|对于大多数使用原始数据的用例，你可以很好地使用原生功能，它做的正是你想要实现的：|1248374|let+uniqueArr+=+[];

uniqueArr+=+[...new+Set+([...array1,+...array2])].sort();|1248375|但我想这不是你的问题。如果数据很复杂，或者只是为了好玩，我们可以考虑其他解决方案。|1248376|做事情的方法有很多种:有些更耗时，有些需要更多的空间。这些是我对这个解决方案的想法，请随意更正或表扬。|1248377|function+binarySearch(arr,+left,+right,+x){
++++//+fastest+search+in+a+sorted+(!)+array
++++
++++if+(right+>=+l)+{
++++++++let+mid+=+left+%2B+Math.floor((right+-+l)+/+2);
++
++++++++if+(arr[mid]+==+x)
++++++++++++return+mid;
++
++++++++if+(arr[mid]+>+x)
++++++++++++return+binarySearch(arr,+left,+mid+-+1,+x);

++++++++return+binarySearch(arr,+mid+%2B+1,+right,+x);
++++}
++
++++//+element+is+not+present+in+array
++++return+-1;
}

function+cloneArrayKeepUniqueOnly(array,+array2){
++++//+array+must+be+sorted,+arg+remains+unmodified
++++//+keeps+only+unique+elements+from+the+array
++++let+uniqueOnly+=+[];

++++for+(let+i+=+0;+i+<+array.length;+i%2B%2B+){
++++++++
++++++++if+(array[i]+==+array[i%2B1]+){
++++++++++++i+%2B%2B;
++++++++++++continue;+
++++++++}+else+if+(array[i]+!=+array[i-1]+&&+binarySearch(array2,+0,+array2.length,+array[i])+==+-1){
++++++++++++
++++++++++++uniqueOnly.push(array[i]);++++++++++
++++++++}
++++}
++++
++++return+uniqueOnly;
}

function+cloneArrKeepOneOfEach(array){
++++//+array+must+be+sorted,+arg+remains+unmodified
++++//+keeps+one+of+every+element+from+the+array
++++let+oneOfEach+=+[];
++++
++++for+(let+i+=+0;+i+<+array.length;+i%2B%2B+){
++++++++
++++++++if+(array[i]+==+array[i%2B1]+){

++++++++++++continue;+
++++++++}+else{
++++++++++++oneOfEach.push(array[i]);
++++++++}
++++}
++++
++++return+oneOfEach;
}

//+main+---------------

let+array1+=+[0,0,0,0,1,1,2,3,4,5,5,5,5,6,7];
let+array2+=+[1,1,1,1,1,1,1,1];

let+mergedArray_oneOfEach+=+[];
let+mergedArray_onlyUniques+=+[];

let+complementArr_uniques,+complementArr;+

//+uniques+only+---------------+start
mergedArray_onlyUniques+=+cloneArrayKeepUniqueOnly(array1,+array2);
complementArr_uniques+=+cloneArrayKeepUniqueOnly(array2,+array1);+

for+(element+of+complementArr_uniques){
++++mergedArray_onlyUniques.push(element);
}

console.log("mergedArray_onlyUniques:+",+mergedArray_onlyUniques);
//+uniques+only+---------------+end

//one+of+each+----------------start

mergedArray_oneOfEach+=+cloneArrKeepOneOfEach(array1);
complementArr+=+cloneArrKeepOneOfEach(array2);

while+(i+<+mergedArray_oneOfEach.length){

++++let+searchIndex+=+binarySearch(complementArr,+0,+complementArr.length-1,+mergedArray_oneOfEach[i]);

++++if+(searchIndex+!=+-1)+{

++++++++complementArr.splice(searchIndex,+1);
++++}+
++++i+%2B%2B;
}

for+(element+of+complementArr){
++++mergedArray_oneOfEach.push(element);
}

console.log("mergedArray_oneOfEach:+",+mergedArray_oneOfEach);

//one+of+each+----------------end|1248378|或者，先合并两个数组，然后对em进行排序，然后使用“缩减/克隆”函数，或者使用哈希表。这将取决于您可以使用的数据和资源的特性；|1248379|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

First of all, your code does not do, what you think it does. Try to run it with this case:
<pre><code>array1 = [0,0,0,1,1,1,3,5,5,5,5,9];
array2 = [1,1,1,1,1,1];
</code></pre>
and you'll see what's wrong.
For most usecases with primitive data, you can go very well with native functionality, it does exactly what you are trying to achieve:
<pre><code>let uniqueArr = [];

uniqueArr = [...new Set ([...array1, ...array2])].sort();
</code></pre>
But this wasn't your question, I assume.
If the data is complex or just for fun we can think of other solutions.
There are many ways to do things: some are more time-intensive, some need more space. These are my ideas for this solution, feel free to correct or credit.
<pre><code>function binarySearch(arr, left, right, x){
 // fastest search in a sorted (!) array
 
 if (right &gt;= l) {
 let mid = left + Math.floor((right - l) / 2);
 
 if (arr[mid] == x)
 return mid;
 
 if (arr[mid] &gt; x)
 return binarySearch(arr, left, mid - 1, x);

 return binarySearch(arr, mid + 1, right, x);
 }
 
 // element is not present in array
 return -1;
}

function cloneArrayKeepUniqueOnly(array, array2){
 // array must be sorted, arg remains unmodified
 // keeps only unique elements from the array
 let uniqueOnly = [];

 for (let i = 0; i &lt; array.length; i++ ){
 
 if (array[i] == array[i+1] ){
 i ++;
 continue; 
 } else if (array[i] != array[i-1] &amp;&amp; binarySearch(array2, 0, array2.length, array[i]) == -1){
 
 uniqueOnly.push(array[i]); 
 }
 }
 
 return uniqueOnly;
}

function cloneArrKeepOneOfEach(array){
 // array must be sorted, arg remains unmodified
 // keeps one of every element from the array
 let oneOfEach = [];
 
 for (let i = 0; i &lt; array.length; i++ ){
 
 if (array[i] == array[i+1] ){

 continue; 
 } else{
 oneOfEach.push(array[i]);
 }
 }
 
 return oneOfEach;
}

// main ---------------

let array1 = [0,0,0,0,1,1,2,3,4,5,5,5,5,6,7];
let array2 = [1,1,1,1,1,1,1,1];

let mergedArray_oneOfEach = [];
let mergedArray_onlyUniques = [];

let complementArr_uniques, complementArr; 

// uniques only --------------- start
mergedArray_onlyUniques = cloneArrayKeepUniqueOnly(array1, array2);
complementArr_uniques = cloneArrayKeepUniqueOnly(array2, array1); 

for (element of complementArr_uniques){
 mergedArray_onlyUniques.push(element);
}

console.log(&quot;mergedArray_onlyUniques: &quot;, mergedArray_onlyUniques);
// uniques only --------------- end

//one of each ----------------start

mergedArray_oneOfEach = cloneArrKeepOneOfEach(array1);
complementArr = cloneArrKeepOneOfEach(array2);

while (i &lt; mergedArray_oneOfEach.length){

 let searchIndex = binarySearch(complementArr, 0, complementArr.length-1, mergedArray_oneOfEach[i]);

 if (searchIndex != -1) {

 complementArr.splice(searchIndex, 1);
 } 
 i ++;
}

for (element of complementArr){
 mergedArray_oneOfEach.push(element);
}

console.log(&quot;mergedArray_oneOfEach: &quot;, mergedArray_oneOfEach);

//one of each ----------------end
</code></pre>
Alternative, merge the two arrays first, than sort em, than use a &quot;reducing/cloning&quot; function, or use a hash table. It would depend on characteristics of data and ressources at your disposal;

I came up with this in order to merge two sorted arrays and remove duplicates. I want to know if I can do better.

Any ideas?

<pre><code>var mergeArrays = function(arr1 , arr2){
var mergedArray = new Array();
var i = 0, j=0,k=0;
var prev = -1;
while(arr1.length &gt; i || arr2.length &gt; j){
 if(arr1[i] == arr2[j]){
 mergedArray[k] = arr1[i];
 i++;
 j++;
 }else if(arr1[i] &lt; arr2[j] || arr2[j] == undefined){
 mergedArray[k] = arr1[i];
 i++;
 }else{
 if(arr2[j]&gt;prev) {
 mergedArray[k] = arr2[j];
 }
 j++;
 }

 prev = mergedArray[k];
 k++;

}

return mergedArray;
}
</code></pre>

Merge sorted arrays and remove duplicates 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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我想出这个方法是为了合并两个排序的数组并删除重复项。我想知道我是否能做得更好。有什么想法吗？var mergeArrays = function(arr1 , arr2){var mergedArray = new Array();var i = 0, j=0,k=0;var prev = -1;while(arr1.length  > i || arr2.length > j){    if(a

问合并已排序的数组并删除重复的javascript
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回答 5

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