python 中递归算法求数组的排列组合

给定数组 nums = [1, 2, 3, 4]，其排列组合共有 种，这里我们需要列出所有的排列组合。

采用递归法进行求：

nums = [1, 2, 3, 4]

def recursion(num):
    
    if len(num) == 1:
        return [num]

    d = []
    for i in range(len(num)):   # 这里不懂得话可以从 nums = [1, 2] 开始
        for j in recursion(num[:i]+num[i+1:]):
            d.append([num[i]] + j)
            
    return d

print(recursion(nums))

输出：

[[1, 2, 3, 4], 
 [1, 2, 4, 3],
 [1, 3, 2, 4],
 [1, 3, 4, 2],
 [1, 4, 2, 3],
 [1, 4, 3, 2],
 [2, 1, 3, 4],
 [2, 1, 4, 3],
 [2, 3, 1, 4],
 [2, 3, 4, 1],
 [2, 4, 1, 3],
 [2, 4, 3, 1],
 [3, 1, 2, 4],
 [3, 1, 4, 2],
 [3, 2, 1, 4],
 [3, 2, 4, 1],
 [3, 4, 1, 2],
 [3, 4, 2, 1],
 [4, 1, 2, 3],
 [4, 1, 3, 2],
 [4, 2, 1, 3],
 [4, 2, 3, 1],
 [4, 3, 1, 2],
 [4, 3, 2, 1]]