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PHP丢框,即:多个数组任意排列组合为一个大数组,同时必须包含所有原数组

时间:2021-07-01 10:21:17 帮助过:17人阅读

php 数组 任意 排列 组合

和部门同时讨论了一上午了,没有讨论出结果来,基本需求如下:
假设:
$a = array('a');
$b = array('b');
$c = array('c');
$d = array('d');

通过一个函数获取以下结果:
array(
a+b+c+d,
ab+c+d,
ac+b+d,
ad+b+c,
bc+ad,
bc+a+d,
bd+ac,
bd+a+c
cd+ab,
cd+a+b,
abc+a,
acd+b,
abd+c,
bcd+a,
...
)

即获取所有数组的组合方式,任意组合数组排序不一定按照顺序,但是要求组合必须出现全部元素。
有高人给个好的算法或者思路么?


回复讨论(解决方案)

用现成的函数实现,未作优化

$a = array('a', 'b', 'c', 'd');$res = array();foreach(arrangement($a) as $v) {  $res = array_merge($res, foo(explode(' ', trim($v))));}print_r($res);//排列 arrangementfunction arrangement($arr = array(), $res = '') {  if(! is_array($arr) ) $arr = str_split($arr);  if(empty($arr)) $array[] = $res;  else foreach($arr AS $k => $v) {    unset($arr[$k]);    foreach( Arrangement($arr, $res . " $v") AS $t) $array[] = $t;    $arr[$k]    = $v;  }  return  $array;}function foo($ar) {  $res = array();  if(count($ar) > 2) {    $t = array_shift($ar);    foreach(foo($ar) as $v) {      $res[] = "$t$v";       $res[] = "$t+$v";    }  }else {    $res[] = "$ar[0]$ar[1]";     $res[] = "$ar[0]+$ar[1]";  }  return $res;}
Array(    [0] => abcd    [1] => a+bcd    [2] => ab+cd    [3] => a+b+cd    [4] => abc+d    [5] => a+bc+d    [6] => ab+c+d    [7] => a+b+c+d    [8] => abdc    [9] => a+bdc    [10] => ab+dc    [11] => a+b+dc    [12] => abd+c    [13] => a+bd+c    [14] => ab+d+c    [15] => a+b+d+c    [16] => acdb    [17] => a+cdb    [18] => ac+db    [19] => a+c+db    [20] => acd+b    [21] => a+cd+b    [22] => ac+d+b    [23] => a+c+d+b    [24] => acbd    [25] => a+cbd    [26] => ac+bd    [27] => a+c+bd    [28] => acb+d    [29] => a+cb+d    [30] => ac+b+d    [31] => a+c+b+d    [32] => adbc    [33] => a+dbc    [34] => ad+bc    [35] => a+d+bc    [36] => adb+c    [37] => a+db+c    [38] => ad+b+c    [39] => a+d+b+c    [40] => adcb    [41] => a+dcb    [42] => ad+cb    [43] => a+d+cb    [44] => adc+b    [45] => a+dc+b    [46] => ad+c+b    [47] => a+d+c+b    [48] => bcda    [49] => b+cda    [50] => bc+da    [51] => b+c+da    [52] => bcd+a    [53] => b+cd+a    [54] => bc+d+a    [55] => b+c+d+a    [56] => bcad    [57] => b+cad    [58] => bc+ad    [59] => b+c+ad    [60] => bca+d    [61] => b+ca+d    [62] => bc+a+d    [63] => b+c+a+d    [64] => bdac    [65] => b+dac    [66] => bd+ac    [67] => b+d+ac    [68] => bda+c    [69] => b+da+c    [70] => bd+a+c    [71] => b+d+a+c    [72] => bdca    [73] => b+dca    [74] => bd+ca    [75] => b+d+ca    [76] => bdc+a    [77] => b+dc+a    [78] => bd+c+a    [79] => b+d+c+a    [80] => bacd    [81] => b+acd    [82] => ba+cd    [83] => b+a+cd    [84] => bac+d    [85] => b+ac+d    [86] => ba+c+d    [87] => b+a+c+d    [88] => badc    [89] => b+adc    [90] => ba+dc    [91] => b+a+dc    [92] => bad+c    [93] => b+ad+c    [94] => ba+d+c    [95] => b+a+d+c    [96] => cdab    [97] => c+dab    [98] => cd+ab    [99] => c+d+ab    [100] => cda+b    [101] => c+da+b    [102] => cd+a+b    [103] => c+d+a+b    [104] => cdba    [105] => c+dba    [106] => cd+ba    [107] => c+d+ba    [108] => cdb+a    [109] => c+db+a    [110] => cd+b+a    [111] => c+d+b+a    [112] => cabd    [113] => c+abd    [114] => ca+bd    [115] => c+a+bd    [116] => cab+d    [117] => c+ab+d    [118] => ca+b+d    [119] => c+a+b+d    [120] => cadb    [121] => c+adb    [122] => ca+db    [123] => c+a+db    [124] => cad+b    [125] => c+ad+b    [126] => ca+d+b    [127] => c+a+d+b    [128] => cbda    [129] => c+bda    [130] => cb+da    [131] => c+b+da    [132] => cbd+a    [133] => c+bd+a    [134] => cb+d+a    [135] => c+b+d+a    [136] => cbad    [137] => c+bad    [138] => cb+ad    [139] => c+b+ad    [140] => cba+d    [141] => c+ba+d    [142] => cb+a+d    [143] => c+b+a+d    [144] => dabc    [145] => d+abc    [146] => da+bc    [147] => d+a+bc    [148] => dab+c    [149] => d+ab+c    [150] => da+b+c    [151] => d+a+b+c    [152] => dacb    [153] => d+acb    [154] => da+cb    [155] => d+a+cb    [156] => dac+b    [157] => d+ac+b    [158] => da+c+b    [159] => d+a+c+b    [160] => dbca    [161] => d+bca    [162] => db+ca    [163] => d+b+ca    [164] => dbc+a    [165] => d+bc+a    [166] => db+c+a    [167] => d+b+c+a    [168] => dbac    [169] => d+bac    [170] => db+ac    [171] => d+b+ac    [172] => dba+c    [173] => d+ba+c    [174] => db+a+c    [175] => d+b+a+c    [176] => dcab    [177] => d+cab    [178] => dc+ab    [179] => d+c+ab    [180] => dca+b    [181] => d+ca+b    [182] => dc+a+b    [183] => d+c+a+b    [184] => dcba    [185] => d+cba    [186] => dc+ba    [187] => d+c+ba    [188] => dcb+a    [189] => d+cb+a    [190] => dc+b+a    [191] => d+c+b+a)

多谢版主,函数我有话下,去重就OK啦。
去重后再次贴出,多谢先,稍后确认。

有重复的吗?显然没有
print_r(array_unique($res));

print_r($res);
的结果一样!

var_dump(count($res) == count(array_unique($res)));
bool(true)

抱歉版主,我没有解释清楚,A+B+C+D 和 B+A+C+D, D+C+A+B等算一种。

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