这是我的代码,以获得所有可能性:
$seq[1] = 'd';
$seq[2] = 'f';
$seq[3] = 'w';
$seq[4] = 's';
for($i = 1; $i < 5; $i++)
{
$s['length_1'][] = $seq[$i];
$c1++;
for($i2 = $i+1; $i2 < 5; $i2++)
{
$s['length_2'][] = $seq[$i].$seq[$i2];
$last = $seq[$i].$seq[$i2];
$c2++;
for($i3 = $i2+1; $i3 < 5; $i3++)
{
$s['length_3'][] = $last.$seq[$i3];
$last = $last.$seq[$i3];
$c3++;
for($i4 = $i3+1; $i4 < 5; $i4++)
{
$s['length_4'][] = $last.$seq[$i4];
$c4++;
}
}
}
}
for($i = 0; $i < $c1; $i++)
echo $s['length_1'][$i].'<br>';
for($i = 0; $i < $c2; $i++)
echo $s['length_2'][$i].'<br>';
for($i = 0; $i < $c3; $i++)
echo $s['length_3'][$i].'<br>';
for($i = 0; $i < $c4; $i++)
echo $s['length_4'][$i].'<br>';
但如果我想添加更多,那么我将不得不再添加一个循环。那么,我怎么能用递归来做呢?我尝试,我尝试,但我真的不能这样做。 请尽可能简单地帮助和发布示例。
谢谢。
答案 0 :(得分:21)
这里有一个算法,
function getCombinations($base,$n){
$baselen = count($base);
if($baselen == 0){
return;
}
if($n == 1){
$return = array();
foreach($base as $b){
$return[] = array($b);
}
return $return;
}else{
//get one level lower combinations
$oneLevelLower = getCombinations($base,$n-1);
//for every one level lower combinations add one element to them that the last element of a combination is preceeded by the element which follows it in base array if there is none, does not add
$newCombs = array();
foreach($oneLevelLower as $oll){
$lastEl = $oll[$n-2];
$found = false;
foreach($base as $key => $b){
if($b == $lastEl){
$found = true;
continue;
//last element found
}
if($found == true){
//add to combinations with last element
if($key < $baselen){
$tmp = $oll;
$newCombination = array_slice($tmp,0);
$newCombination[]=$b;
$newCombs[] = array_slice($newCombination,0);
}
}
}
}
}
return $newCombs;
}
我知道它在任何方面都没有效果,但在小套装中使用应该不是问题
first base参数是一个数组,包含生成组合时要考虑的元素。
简单用法和输出:
var_dump(getCombinations(array("a","b","c","d"),2));
,输出
array
0 =>
array
0 => string 'a' (length=1)
1 => string 'b' (length=1)
1 =>
array
0 => string 'a' (length=1)
1 => string 'c' (length=1)
2 =>
array
0 => string 'a' (length=1)
1 => string 'd' (length=1)
3 =>
array
0 => string 'b' (length=1)
1 => string 'c' (length=1)
4 =>
array
0 => string 'b' (length=1)
1 => string 'd' (length=1)
5 =>
array
0 => string 'c' (length=1)
1 => string 'd' (length=1)
要列出数组的所有子集,请使用此组合算法执行
$base =array("a","b","c","d");
for($i = 1; $i<=4 ;$i++){
$comb = getCombinations($base,$i);
foreach($comb as $c){
echo implode(",",$c)."<br />";
}
}
输出
a
b
c
d
a,b
a,c
a,d
b,c
b,d
c,d
a,b,c
a,b,d
a,c,d
b,c,d
a,b,c,d
答案 1 :(得分:14)
这是一个简单的算法。从1到2 count(array) -1迭代。在每次迭代中,如果循环计数器的二进制表示中的第j位等于1,则在组合中包含第j个元素。
由于PHP需要能够将2 count(array)计算为整数,因此这可能永远不会超过PHP_INT_MAX
。在64位PHP安装中,您的数组不能超过62个元素,因为2 62 保持低于PHP_INT_MAX
而2 63 超过它。
编辑:这计算所有可能的组合,而不是排列(即'abc'='cba')。它通过用二进制表示原始数组并从0“向上计数”到完整数组的二进制表示来实现,有效地构建了每个可能的唯一组合的列表。
$a = array('a', 'b', 'c', 'd');
$len = count($a);
$list = array();
for($i = 1; $i < (1 << $len); $i++) {
$c = '';
for($j = 0; $j < $len; $j++)
if($i & (1 << $j))
$c .= $a[$j];
$list[] = $c;
}
print_r($list);
答案 2 :(得分:3)
这是:
<?php
function combinations($text,$space)
{
// $text is a variable which will contain all the characters/words of which we want to make all the possible combinations
// Let's make an array which will contain all the characters
$characters=explode(",", $text);
$x=count($characters);
$comb = fact($x);
// In this loop we will be creating all the possible combinations of the positions that are there in the array $characters
for ($y=1; $y<= $comb; $y++)
{
$ken = $y-1;
$f = 1;
$a = array();
for($iaz=1; $iaz<=$x; $iaz++)
{
$a[$iaz] = $iaz;
$f = $f*$iaz;
}
for($iaz=1; $iaz<=$x-1; $iaz++)
{
$f = $f/($x+1-$iaz);
$selnum = $iaz+$ken/$f;
$temp = $a[$selnum];
for($jin=$selnum; $jin>=$iaz+1; $jin--)
{
$a[$jin] = $a[$jin-1];
}
$a[$iaz] = $temp;
$ken = $ken%$f;
}
$t=1;
// Let’s start creating a word combination: we have all the necessary positions
$newtext="";
// Here is the while loop that creates the word combination
while ($t<=$x)
{
$newtext.=$characters[$a[$t]-1]."$space";
$t++;
}
$combinations[] = $newtext ;
}
return $combinations;
}
function fact($a){
if ($a==0) return 1;
else return $fact = $a * fact($a-1);
}
$a = combinations("d,f,w,s","");
foreach ($a as $v) {
echo "$v"."\n";
}
?>
<强>输出:强>
dfws
dfsw
dwfs
dwsf
dsfw
dswf
fdws
fdsw
fwds
fwsd
fsdw
fswd
wdfs
wdsf
wfds
wfsd
wsdf
wsfd
sdfw
sdwf
sfdw
sfwd
swdf
swfd
此外,read this;
答案 3 :(得分:1)
你可以这样做:
function combinations($arr) {
$combinations = array_fill(0, count($arr)+1, array());
$combinations[0] = array('');
for ($i = 0, $n = count($arr); $i < $n; ++$i) {
for ($l = $n-$i; $l > 0; --$l) {
$combinations[$l][] = implode('', array_slice($arr, $i, $l));
}
}
return $combinations;
}
以下是一个例子:
$arr = array('d', 'f', 'w', 's');
var_dump(combinations($arr));
这会生成以下数组:
array(
array(''), // length=0
array('d', 'f', 'w', 's'), // length=1
array('df', 'fw', 'ws'), // length=2
array('dfw', 'fws'), // length=3
array('dfws') // length=4
)
简要说明:
对于每个 i ,其中0≤ i &lt; n ,获取所有子阵列 arr [ i , i + l ],每个可能的长度为0&lt; l ≤ n - i 。
答案 4 :(得分:0)
这是一个标准的排列问题,如果您需要字符串的所有变体,请查看“php permutations”。
答案 5 :(得分:0)
这是我打印所有可能的字符组合的功能:
function printCombinations($var, $begin = 0, $preText = "") {
for($i = $begin; $i < count($var); $i++) {
echo $preText . $var[$i] . "\n";
if(($i+1) < count($var))
printCombinations($var, $i+1, $preText . $var[$i]);
}
}
printCombinations(array('a','b','c','d','e'));