向递归函数传递参数的不同类型

Question

下面的代码工作得很好。

        def perm(cur):
            avail = [i for i in nums if not i in cur]
            if len(avail) == 0:
                res.append(cur.copy())
                return
            
            for i in avail:
                cur.append(i)
                perm(cur)
                cur.pop()
        
        perm([])
        return res

然而，下面的返回错误

         def perm(cur):
            avail = [i for i in nums if not i in cur]
            if len(avail) == 0:
                res.append(cur.copy())
                return
            
            for i in avail:
                perm(cur.append(i))
                cur.pop()
        
        perm([])
        return res

TypeError: argument of type 'NoneType' is not iterable
    avail = [i for i in nums if not i in cur]
Line 5 in <listcomp> (Solution.py)
    avail = [i for i in nums if not i in cur]
Line 5 in perm (Solution.py)
    perm(cur.append(i))
Line 12 in perm (Solution.py)
    perm([])
Line 15 in permute (Solution.py)
    ret = Solution().permute(param_1)
Line 38 in _driver (Solution.py)
    _driver()
Line 49 in <module> (Solution.py)

我正在尝试理解这背后的逻辑。

我不确定这两种参数传递方式之间的区别。

Answer 1

.append returns None.

def perm(cur):
        avail = [i for i in nums if not i in cur]
        if len(avail) == 0:
            res.append(cur.copy())
            return
        
        for i in avail:
            // Here .append(i) does not return anything so the cur you are passing in the perm funcion is None
            perm(cur.append(i))
            cur.pop()
    
    perm([])
    return res

Answer 2

函数 cur.append(i) 没有 return 任何东西。它仅将 i 附加到 cur 列表。

然后当你用它作为参数调用 perm 时，你基本上调用了 perm(None)，这是不可迭代的。因此 [i for i in nums if not i in cur] 失败。

向递归函数传递参数的不同类型

Different types of passing parameters to a recursive function

recursion