程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
您现在的位置: 程式師世界 >> 編程語言 >  >> 更多編程語言 >> Python

Python description leetcode 78 subset

編輯:Python

Python describe LeetCode 78. A subset of

Hello everyone , I'm Qi Guanjie (qí guān jié ), stay 【 Qi Guanjie 】 official account 、CSDN、GitHub、B Share some technical blog posts on the website and other platforms , It mainly includes front-end development 、python The backend development 、 Applet development 、 Data structure and algorithm 、docker、Linux Common operation and maintenance 、NLP And other related technical blog , Time flies , Future period , come on. ~

If you love bloggers, you can focus on the bloggers' official account. 【 Qi Guanjie 】(qí guān jié), The articles inside are more complete and updated faster . If you need to find a blogger, you can leave a message at the official account. , I will reply to the message as soon as possible .


This article was originally written as 【 Qi Guanjie 】(qí guān jié ), Please support the original , Some platforms have been stealing blog articles maliciously !!! All articles please pay attention to WeChat official account 【 Qi Guanjie 】.

subject

Give you an array of integers nums , Elements in an array Different from each other . Returns all possible subsets of the array ( Power set ).

Solution set You can't Contains a subset of repetitions . You can press In any order Returns the solution set .

Example 1:

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

Example 2:

 Input :nums = [0]
Output :[[],[0]]

Tips :

  • 1 <= nums.length <= 10
  • -10 <= nums[i] <= 10
  • nums All elements in Different from each other

Python describe

Explosive search

class Solution:
def subsets(self, nums: List[int]) -> List[List[int]]:
res = []
tmp = []
n = len(nums)
def dfs(idx):
nonlocal n
if idx == n:
res.append(copy.deepcopy(tmp))
return
dfs(idx+1)
tmp.append(nums[idx])
dfs(idx+1)
tmp.pop()
dfs(0)
return res

  1. 上一篇文章:
  2. 下一篇文章:
Copyright © 程式師世界 All Rights Reserved