blocks|key|1573973|text|如果你需要10的任意更高的倍数，最简单的方法就是乘以10。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1573974|const+solution+=+n+=>+n+*+10|code-block|syntax|javascript|1573975|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>If you need arbitrary higher number multiple of 10.
the easiest approach is to multiply it by 10.</p>

<pre><code>const solution = n =&gt; n * 10
</code></pre>


blocks|key|1789174|text|您可以只划分并使用Math.ceil|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1789175|​|1789176|const+solution+=+(n)+=>+Math.ceil((n%2B1)/10)+*+10+

console.log(solution(49))
console.log(solution(40))
console.log(solution(41))
console.log(solution(0))
console.log(solution(-1))
console.log(solution(-11))|code-block|syntax|javascript|1789177|1789178|entityMap^0|9|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|T|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

<p>You could just divide and use <code>Math.ceil</code>:</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 solution = (n) =&gt; Math.ceil((n+1)/10) * 10 

console.log(solution(49))
console.log(solution(40))
console.log(solution(41))
console.log(solution(0))
console.log(solution(-1))
console.log(solution(-11))</code></pre>
</div>
</div>
</p>


blocks|key|1574036|text|没有数学，使用mod。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1574037|​|1574038|const+solution+=+n+=>+n+%2B+10+-+n+%25+10

console.log(solution(57))
console.log(solution(23))
console.log(solution(1))
console.log(solution(221))
console.log(solution(9000))|code-block|syntax|javascript|1574039|1574040|对于小于10%5E9的N，您总是可以添加一个if来捕获这些数字。|1574041|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|R|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|M|$]]

<p>Without Math, using mod.</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 solution = n =&gt; n + 10 - n % 10

console.log(solution(57))
console.log(solution(23))
console.log(solution(1))
console.log(solution(221))
console.log(solution(9000))</code></pre>
</div>
</div>
</p>

<p>For N being smaller than 10^9 you can always add an if to catch those numbers.</p>


blocks|key|1574181|text|不确定您是否需要处理负数，但您可以使用我想到的这个技巧，它将给您提供大于n的下一个数字，该数字以0结尾：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1574182|​|1574183|function+solution(number)+{
	var+remainder+=+number+%25+10;
	if(remainder+===+0)+{
		return+number+%2B+10;
	}+else+{
		return+number+%2B+(10+-+remainder);
	}
}

console.log(solution(42));
console.log(solution(50));
console.log(solution(105));|code-block|syntax|javascript|1574184|1574185|编辑:+solution+by+assoron使用相同的逻辑，但只有一条线。|1574186|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|R|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|M|$]]

<p>Not sure if you need to work with negative numbers, but you can use this trick I thought of which will give you the next number higher than n which ends in 0:</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 solution(number) {
	var remainder = number % 10;
	if(remainder === 0) {
		return number + 10;
	} else {
		return number + (10 - remainder);
	}
}

console.log(solution(42));
console.log(solution(50));
console.log(solution(105));</code></pre>
</div>
</div>
</p>

<p>Edit: solution by assoron uses the same logic, but is a one liner.</p>


blocks|key|1574212|text|由于没有人考虑精度问题，因此合理的假设是结果必须在安全整数内，所以我们可以简单地返回最大安全输入：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1574213|const+safe+=+Number.MAX_SAFE_INTEGER-1
const+solution+=+n+=>+safe
console.log(solution(2))|code-block|syntax|javascript|1574214|注意，结果是预先计算的，所以我们只需要从内存中加载一个常量并返回它。|1574215|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>Since noone have considered the problem of precision, then it is a reasonable assumption that the result must be inside the safe integers, so we can simply return the maximal safe enter:</p>

<pre><code>const safe = Number.MAX_SAFE_INTEGER-1
const solution = n =&gt; safe
console.log(solution(2))
</code></pre>

<p>Note that the result is precomputed, so we only need to load a constant from memory and return it.</p>


blocks|key|1574260|text|function+solution(n)
{
++++return+n+%25+10+?+n+%2B+(10+-+n+%25+10)+:+n+%2B+10;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1574261|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 solution(n)
{
    return n % 10 ? n + (10 - n % 10) : n + 10;
}
</code></pre>


<p>Had a coding challenge to do.</p>

<p>The question was to write a function which returns an arbitrary higher number than N which ends in 0. </p>

<p>My solution was:</p>

<pre><code>function solution(N) {
    for (let n = N +1; n &lt;= N + 10; n++) {
        if (n % 10 === 0) return n;
    }
}
</code></pre>


What would be a more performant way of solving this problem with JavaScript

有一个编码挑战要做。问题是要写一个函数，它返回一个比N大的任意数字，并以0结尾。我的解决方案是：function solution(N) {    for (let n = N +1; n <= N + 10; n++) {        if (n % 10 === 0) return n;    }}

问使用JavaScript解决此问题的更有效的方法是什么
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