由网友(十足痞子味)分享简介：任何人都可以提出让所有的组合，而无需重复一个简单的方法，并用不同长度？Can anyone suggest an easy way of getting all combinations without repeats, and with a varying length ?[0,1,2,3,4] [0,1,2,3...

任何人都可以提出让所有的组合，而无需重复一个简单的方法，并用不同长度？

Can anyone suggest an easy way of getting all combinations without repeats, and with a varying length ?

[0,1,2,3,4]

[0,1,2,3,4]

（2的）： [0,1]，[0,2]，[0,3] ...... [3,4]

(2's) : [0,1],[0,2],[0,3] ... [3,4]

（3的）： [0,1,2]，[0,1,3] ... [2,3,4]

(3's) : [0,1,2],[0,1,3] ... [2,3,4]

（4的）： [0,1,2,3]，[0,1,2,4] ... [1,2,3,4]

(4's) : [0,1,2,3],[0,1,2,4] ... [1,2,3,4]

推荐答案

我花了一段时间，但我想我已经在这里了... ...

took me a while but I think I've got it here...

这所有的组合，而无需重复

this has all the combinations without repeats

var arr:Array = [0, 1, 2, 3, 4];
var $allcombos:Array = [];
findCombos($allcombos,[],arr);

function findCombos($root:Array, $base:Array, $rem:Array):void {
    for (var i:int = 0; i < $rem.length; i++) {
        var a:Array = $base.concat();
        a.push($rem[i]);
        findCombos($root, a, $rem.slice(i + 1));
        if (a.length > 1) {
            $root.push(a);
        }
    }
}