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需要帮助了解递归函数以生成字符串的所有组合

[英]Need help understanding recursive function to generate all combinations of a string

I came across this code , but am having a hard time figuring out how the flow of this function is working. 我遇到了这段代码 ,但我很难弄清楚这个函数的流程是如何工作的。

function combinations(str) {
    var fn = function(active, rest, a) {
        if (!active && !rest)
            return;
        if (!rest) {
            a.push(active);
        } else {
            fn(active + rest[0], rest.slice(1), a);
            fn(active, rest.slice(1), a);
        }
        return a;
    }
    return fn("", str, []);
}

I've console.logged a bunch of statements, but I'm getting lost in the recursion. 我控制台。记录了一堆语句,但我在递归时迷路了。 Specifically, I don't understand what's happening after the first of the two fn 's gets returned. 具体来说,我不明白在两个fn的第一个返回后发生了什么。 It's simply returning a , but to me, it appears as if the second fn knows to take the returned a . 它只是返回a ,但对我来说,似乎第二个fn知道fna How does that work? 这是如何运作的? Shouldn't it need a variable assignment like so: 它不应该像这样需要变量赋值:

a = fn(active + rest[0], rest.slice(1), a);
fn(active, rest.slice(1), a);

Well, if you're trying to understand the specific algorithm, the key insight is the a variable accumulates the result by reference -- the lower level calls to combinations() push individual elements onto a and the function eventually returns accumulated result. 好吧,如果你想了解具体的算法,关键洞察力是a变量引用积累的结果-下级调用combinations()推动各个元素到a与函数最终返回积累的结果。

A good print statement can help understanding cases like this. 一个好的打印声明可以帮助理解这样的情况。 Here's the code with the addition of a print out that shows you what level of recursion you're at. 这是添加了打印输出的代码,显示了您所处的递归级别。 And I renamed a to accum to make it a bit more obvious: 我将a重命名为accum以使其更明显:

function combinations(str) {
  var fn = function(active, rest, accum, level) {
    console.log(new Array(level).join('--'),
                ' active: \'' + active + '\'',
                'rest: \'' + rest + '\'',
                'accum:', accum);
    if (!active && !rest)
      return;
    if (!rest) {
      accum.push(active);
    } else {
      fn(active + rest[0], rest.slice(1), accum, level + 1);
      fn(active, rest.slice(1), accum, level + 1);
    }
    return accum;
  }
  return fn("", str, [], 1);
}

and the results: 结果:

combinations('abc')
  active: '' rest: 'abc' accum: []
--  active: 'a' rest: 'bc' accum: []
----  active: 'ab' rest: 'c' accum: []
------  active: 'abc' rest: '' accum: []
------  active: 'ab' rest: '' accum: [ 'abc' ]
----  active: 'a' rest: 'c' accum: [ 'abc', 'ab' ]
------  active: 'ac' rest: '' accum: [ 'abc', 'ab' ]
------  active: 'a' rest: '' accum: [ 'abc', 'ab', 'ac' ]
--  active: '' rest: 'bc' accum: [ 'abc', 'ab', 'ac', 'a' ]
----  active: 'b' rest: 'c' accum: [ 'abc', 'ab', 'ac', 'a' ]
------  active: 'bc' rest: '' accum: [ 'abc', 'ab', 'ac', 'a' ]
------  active: 'b' rest: '' accum: [ 'abc', 'ab', 'ac', 'a', 'bc' ]
----  active: '' rest: 'c' accum: [ 'abc', 'ab', 'ac', 'a', 'bc', 'b' ]
------  active: 'c' rest: '' accum: [ 'abc', 'ab', 'ac', 'a', 'bc', 'b' ]
------  active: '' rest: '' accum: [ 'abc', 'ab', 'ac', 'a', 'bc', 'b', 'c' ]
[ 'abc', 'ab', 'ac', 'a', 'bc', 'b', 'c' ]

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