Leetcode491 increasing subsequence

Give you an array of integers nums , Find and return all different increasing subsequences in the array , In increasing subsequence There are at least two elements . You can press In any order Return to the answer .

The array may contain duplicate elements , If two integers are equal , It can also be regarded as a special case of increasing sequence .

Example 1：

Input ：nums = [4,6,7,7]
Output ：[[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]
Example 2：

Input ：nums = [4,4,3,2,1]
Output ：[[4,4]]

class Solution:
    def findSubsequences(self, nums: List[int]) -> List[List[int]]:
            result = list()
            self.dfs(nums, result, list(), 0)
            return result

    def dfs(self, nums, result, path, index):
        n = len(nums)
        #  Satisfied handle path When adding a condition to the result , You can't return, because path You can continue to add 
        if len(path) >= 2:
            result.append(path.copy())
        if index >= n:
            return

        used_set = set()
        for i in range(index, n):
            #  Generally, when there are repeated elements , It only needs nums[i]!=nums[i-1] To and fro 
            #  The premise of this method is that the array implementation has been sorted ,
            #  So as long as the repeated items in the array will be next to each other .
            #  But here is the subsequence problem , Reorder not allowed , There can be any number of hops between elements , So as long as you have used it before 
            if nums[i] in used_set:
                continue
            if (len(path) == 0) or (nums[i] >= path[-1]):
                used_set.add(nums[i])
                path.append(nums[i])
                self.dfs(nums, result, path, i + 1)
                path.pop()