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现在有一个数组如下:

<code>$a = [1,2,3,4,5,6,7,8,9]
</code>

现在要通过上面的数据元素重新生成长度为2,3,4...的新数组,并且统计一共组成了新数组的个数.
比如生成长度为2的数字,并且统计$b个数 例:

<code>$b = [1,2]
$b = [1,3]
$b = [1,4]
$b = [1,5]
</code>

现在使用的一个比较笨的方法就是使用循环.长度为2就使用2个循环,3个就用3个循环.这样代码不利于维护.如果有10个,那么要10个循环.

回复内容:

现在有一个数组如下:

<code>$a = [1,2,3,4,5,6,7,8,9]
</code>

现在要通过上面的数据元素重新生成长度为2,3,4...的新数组,并且统计一共组成了新数组的个数.
比如生成长度为2的数字,并且统计$b个数 例:

<code>$b = [1,2]
$b = [1,3]
$b = [1,4]
$b = [1,5]
</code>

现在使用的一个比较笨的方法就是使用循环.长度为2就使用2个循环,3个就用3个循环.这样代码不利于维护.如果有10个,那么要10个循环.

<code>$a = [1,2,3,4,5,6,7,8,9];
// 需要分片的大小
$perSize = 2;

$leader = array_shift($a);
$bs = array_chunk($a, $perSize - 1);
$bs = array_map(function ($item) use ($leader) {
    array_unshift($item, $leader);
    return $item;
}, $bs);
// 所有的 $b, 和 $b 的个数
var_dump($bs, count($bs));
</code>

http://blog.sina.cn/dpool/blog/s/blog_4dfb08c901011wut.html?vt=4

<code>#include <stdio.h>

int array[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int array2[9];
int array2len = 0;

void perm(int len, int index)
{
    int i, j;
    if (array2len </stdio.h></code>

我的思路是用递归,代码如上,C语言的。

对了,要统计的话,有两个思路:一个是在上面的代码中加一些逻辑,在每次需要输出array2的时候统计;另一个是用组合数的计算公式直接计算,比如题中9个数字生成的所有组合共有502个,高中数学讲过。

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