nums = [1, 2, 3]
# 回溯算法，复杂度较高，因为回溯算法就是暴力穷举，遍历整颗决策树是不可避免的
res = []

def backtrack(path=None, selects=None):
    if path is None:
        path = []  # 用来存放符合条件的结果
    if not selects:
        res.append(path[:])  # 此时说明找到了一组
        return
    for i in range(len(selects)):
        path.append(selects[i])
        backtrack(path, selects[:i] + selects[i + 1:])  # 递归
        path.pop()  # 回溯


backtrack(selects=nums)
print(res)