blocks|key|4094376|text|对于10个数字，最简单的方法是遍历所有组合并计算差值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4094377|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

With 10 numbers, the easiest way would be to go over all combinations and calculate the difference.

blocks|key|1485600|text|这类似于Knapsack+problem：您尝试将个人放在其中一个团队中，以便该团队的和是不大于总和的一半的最大值。如果团队规模不受限制，情况也是一样的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1485601|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Knapsack_problem^0|4|G|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

This is similar to the <a href="https://en.wikipedia.org/wiki/Knapsack_problem" rel="nofollow">Knapsack problem</a>: You try to put individuals in one of the teams so that this team's sum is the biggest value not larger than half of the total sum. It would be the same if team size was not restricted.

blocks|key|5562724|text|生成5个元素的所有组合。你将有这5个人在一个团队中，其余的在另一个团队中。比较所有结果并选择差异最小的结果。你可以用5个for循环来创建所有这些组合。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5562725|entityMap^0|1O|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Generate all combination of 5 elements. You will have those 5 in a a team and the remaining in the other team. Compare all results and choose the one with the smallest difference. You can create all those combination with 5 <code>for</code> loops.

blocks|key|1777043|text|我只是尝试了一下--不幸的是，我不得不编写这个置换函数(next函数)，并为每个元素调用result.fit。可以做得更好，但对于演示来说，它应该足够好。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1777044|var+all+=+[+3,+4,+5,+8+,+2,+1,+1,+4,+9,+10+];

function+sumArray(a)+{
++++var+asum+=+0;
++++a.forEach(function(v){+asum+%2B=+v+});
++++return+asum;
}

var+next+=+function(start,+rest,+nbr,+result)+{
++++if+(nbr+<+0)+{
++++++++result.fit(start);
++++++++return;
++++}
++++for+(var+i+=+0;+i+<+rest.length+-+nbr;+%2B%2Bi)+{
++++++++var+clone+=+start.slice(0);
++++++++clone.push(rest[i]);
++++++++next(clone,+rest.slice(i+%2B+1),+nbr+-+1,+result);
++++}
};

var+result+=+{
++++target:+sumArray(all)+/+2,
++++best:+[],
++++bestfit:+Math.pow(2,63),++++//+really+big
++++fit:+function(a)+{
++++++++var+asum+=+sumArray(a);
++++++++var+fit+=+Math.abs(asum+-+this.target);
++++++++if+(fit+<+this.bestfit)+{
++++++++++++this.bestfit+=+fit;
++++++++++++this.best+=+a;
++++++++}
++++}
}

next([],+all,+all.length+/+2,+result);
console.log(JSON.stringify(result.best));|code-block|syntax|javascript|1777045|entityMap^0|S|4|18|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

I just tried it out - unfortunately I had to program that permutation thing (function <code>next</code>) and call <code>result.fit</code> for every element.
Can be done nicer, but for demonstration it should be good enough.

<pre><code>var all = [ 3, 4, 5, 8 , 2, 1, 1, 4, 9, 10 ];

function sumArray(a) {
 var asum = 0;
 a.forEach(function(v){ asum += v });
 return asum;
}

var next = function(start, rest, nbr, result) {
 if (nbr &lt; 0) {
 result.fit(start);
 return;
 }
 for (var i = 0; i &lt; rest.length - nbr; ++i) {
 var clone = start.slice(0);
 clone.push(rest[i]);
 next(clone, rest.slice(i + 1), nbr - 1, result);
 }
};

var result = {
 target: sumArray(all) / 2,
 best: [],
 bestfit: Math.pow(2,63), // really big
 fit: function(a) {
 var asum = sumArray(a);
 var fit = Math.abs(asum - this.target);
 if (fit &lt; this.bestfit) {
 this.bestfit = fit;
 this.best = a;
 }
 }
}

next([], all, all.length / 2, result);
console.log(JSON.stringify(result.best));
</code></pre>

blocks|key|5562794|text|与most相同的算法--比较126个组合。Haskell中的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5562795|inv+=+[1,2,3,4,5,6,7,8,9,10]

best+(x:xs)+(a,b)
++%7C+length+a+==+5+=+[(abs+(sum+a+-+sum+(x:xs+%2B%2B+b)),(a,x:xs+%2B%2B+b))]
++%7C+length+b+==+5+=+[(abs+(sum+(x:xs+%2B%2B+a)+-+sum+b),(x:xs+%2B%2B+a,b))]
++%7C+otherwise+++++=+let+s+=+best+xs+(x:a,b)
++++++++++++++++++++++++s'+=+best+xs+(a,x:b)
++++++++++++++++++++in+if+fst+(head+s)+<+fst+(head+s')+then+s
+++++++++++++++++++++++else+if+fst+(head+s')+<+fst+(head+s)+then+s'
+++++++++++++++++++++++else+s+%2B%2B+s'

main+=+print+$+best+(tail+inv)+([head+inv],[])|code-block|syntax|javascript|5562796|输出：|5562797|*Main>+main
[(1,([9,10,5,2,1],[8,7,6,4,3])),(1,([10,8,6,2,1],[9,7,5,4,3]))
,(1,([9,10,6,2,1],[8,7,5,4,3])),(1,([9,8,7,2,1],[10,6,5,4,3]))
,(1,([10,8,7,2,1],[9,6,5,4,3])),(1,([9,10,4,3,1],[8,7,6,5,2]))
,(1,([10,8,5,3,1],[9,7,6,4,2])),(1,([9,10,5,3,1],[8,7,6,4,2]))
,(1,([10,7,6,3,1],[9,8,5,4,2])),(1,([9,8,6,3,1],[10,7,5,4,2]))
,(1,([10,8,6,3,1],[9,7,5,4,2])),(1,([9,8,7,3,1],[10,6,5,4,2]))
,(1,([10,7,5,4,1],[9,8,6,3,2])),(1,([9,8,5,4,1],[10,7,6,3,2]))
,(1,([10,8,5,4,1],[9,7,6,3,2])),(1,([9,7,6,4,1],[10,8,5,3,2]))
,(1,([10,7,6,4,1],[9,8,5,3,2])),(1,([9,8,6,4,1],[10,7,5,3,2]))
,(1,([8,7,6,5,1],[9,10,4,3,2])),(1,([9,7,6,5,1],[10,8,4,3,2]))]|5562798|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Same algorithm as most -- compare 126 combinations. Code in Haskell:

<pre><code>inv = [1,2,3,4,5,6,7,8,9,10]

best (x:xs) (a,b)
 | length a == 5 = [(abs (sum a - sum (x:xs ++ b)),(a,x:xs ++ b))]
 | length b == 5 = [(abs (sum (x:xs ++ a) - sum b),(x:xs ++ a,b))]
 | otherwise = let s = best xs (x:a,b)
 s' = best xs (a,x:b)
 in if fst (head s) &lt; fst (head s') then s
 else if fst (head s') &lt; fst (head s) then s'
 else s ++ s'

main = print $ best (tail inv) ([head inv],[])
</code></pre>

Output:

<pre><code>*Main&gt; main
[(1,([9,10,5,2,1],[8,7,6,4,3])),(1,([10,8,6,2,1],[9,7,5,4,3]))
,(1,([9,10,6,2,1],[8,7,5,4,3])),(1,([9,8,7,2,1],[10,6,5,4,3]))
,(1,([10,8,7,2,1],[9,6,5,4,3])),(1,([9,10,4,3,1],[8,7,6,5,2]))
,(1,([10,8,5,3,1],[9,7,6,4,2])),(1,([9,10,5,3,1],[8,7,6,4,2]))
,(1,([10,7,6,3,1],[9,8,5,4,2])),(1,([9,8,6,3,1],[10,7,5,4,2]))
,(1,([10,8,6,3,1],[9,7,5,4,2])),(1,([9,8,7,3,1],[10,6,5,4,2]))
,(1,([10,7,5,4,1],[9,8,6,3,2])),(1,([9,8,5,4,1],[10,7,6,3,2]))
,(1,([10,8,5,4,1],[9,7,6,3,2])),(1,([9,7,6,4,1],[10,8,5,3,2]))
,(1,([10,7,6,4,1],[9,8,5,3,2])),(1,([9,8,6,4,1],[10,7,5,3,2]))
,(1,([8,7,6,5,1],[9,10,4,3,2])),(1,([9,7,6,5,1],[10,8,4,3,2]))]
</code></pre>

blocks|key|4094491|text|这是Partition+problem的一个实例，但是对于您的小型实例测试，所有的组合都应该足够快。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4094492|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Partition_problem^0|2|H|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

This is an instance of the <a href="https://en.wikipedia.org/wiki/Partition_problem" rel="nofollow">Partition problem</a>, but for your tiny instance testing all combinations should be fast enough.

So I have 10 numbers. Lets say each number represents the skill of an individual. If I were to create 2 teams of 5 , how would i make 2 teams such that the difference of their teams sum is minimal.

Easy way of bruteforcing numbers

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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功能1上新10个字符

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功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符。

功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符

功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符

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所以我有10个数字。让我们假设每个数字代表一个人的技能。如果我要创建两个5人的团队，我将如何创建两个团队，使他们的团队总和的差异最小。

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