我需要获得任意长度的一系列独特排列:
值数组:a,b,c,d,e,f,g,h,i,j,k,l,m(共13个)
预期结果:a,b,c,ab,ac,bc,abc,....
ab == ba(复制)
abc == acb == bca == bac == ...(重复)
到目前为止我所做的是对它的蛮力攻击,但是有13个元素,这是一种乐观的态度。我需要更聪明的东西,不幸的是我的数学能力之外的东西。
我目前的疯狂解决方案:
// Returns the total number of $count-length strings generatable from $letters.
function getPermCount($letters, $count) {
$result = 1;
// k characters from a set of n has n!/(n-k)! possible combinations
for($i = strlen($letters) - $count + 1; $i <= strlen($letters); $i++) {
$result *= $i;
}
return $result;
}
// Decodes $index to a $count-length string from $letters, no repeat chars.
function getPerm($letters, $count, $index) {
$result = array();
for($i = 0; $i < $count; $i++) {
$pos = $index % strlen($letters);
$result[] = $letters[$pos];
$index = ($index-$pos)/strlen($letters);
$letters = substr($letters, 0, $pos) . substr($letters, $pos+1);
}
sort($result); // to be able and remove dupes later
return implode("", $result);
}
$r = array();
$letters = 'abcdefghijklm';
$len = strlen($letters);
$b = 0;
for ($c = 1; $c <= $len; $c++) {
$p = getPermCount($letters, $c);
echo $c." {".$p." perms} - ";
for($i = 0; $i < $p; $i++) {
$r[] = getPerm($letters, $c, $i)." ";
if ($b > 4000000) {
$r = array_flip($r); // <= faster alternative to array_unique
$r = array_flip($r); // flipping values to keys automaticaly removes dupes
$b = 0;
} else {
$b++;
}
}
$r = array_flip($r); // <= faster alternative to array_unique
$r = array_flip($r); // flipping values to keys automaticaly removes dupes
echo count($r)." unique\n";
}
print_r($r);
答案 0 :(得分:0)
我对它采取了一个裂缝:
function get_perms(chars) {
var perms = [ chars[0] ];
for (var i = 1; i < chars.length; i += 1) {
len = perms.length;
for (var j = 0; j < len; j += 1) {
perms.push(perms[j] + chars[i]);
}
}
return perms;
}
var chars = [ "a", "b", "c", "d", "e" ], perms = [];
for (var i = 0; i < chars.length; i += 1) {
perms = perms.concat(get_perms(chars.slice(i)));
}
console.log(perms);
大纲是:
从完整集[ "a", "b", "c", "d", "e", .. ]
生成所有提升排列,a
,ab
,ac
,abc
等,但绝不会acb
。我这样做是从["a"]
开始,现在将"b"
添加到所有这些并合并以获取:["a", "ab"]
,现在将"c"
添加到所有这些并合并:{{1等等。
为集["a", "ab", "ac", "abc"]
再为[ "b", "c", "d", "e", .. ]
等再次执行此操作。
这应该为您提供所有排列而不重复。
示例输出([ "c", "d", "e", .. ]
):
[ "a", "b", "c", "d", "e" ]
在我的机器上生成一整套13个字符所需的时间不到1秒,因此我不会费心去测量。另外,我意识到我做的是JavaScript而不是PHP,它不应该太难转换。