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17. 电话号码的字母组合 给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。# @Time : 2023/10/13 00:00 # @Author : Lan # @File : 17. 电话号码的字母组合.py # @Software: PyCharm # @link : https://leetcode.cn/problems/letter-combinations-of-a-phone-number/description/ """ 给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。 给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。 """ from typing import List class Solution: key_value = [ "", "", "abc", "def", "ghi", "jkl", "nmo", "pqrs", "tuv", "wxyz" ] result = [] temp = [] def backtracking(self, digits, index): if index == len(digits): if self.temp: self.result.append(''.join(self.temp)) return letters = self.key_value[int(digits[index])] for i in range(len(letters)): self.temp.append(letters[i]) self.backtracking(digits, index + 1) self.temp.pop() def letterCombinations(self, digits: str) -> List[str]: self.result = [] self.backtracking(digits, 0) return self.result print(Solution().letterCombinations(''))
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77. 组合 给定两个整数 n 和 k，返回范围 [1, n] 中所有可能的 k 个数的组合。你可以按 任何顺序 返回答案。示例 1：输入：n = 4, k = 2输出：[ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4],]示例 2：输入：n = 1, k = 1输出：[[1]]提示：1 <= n <= 201 <= k <= n# @Time : 2023/10/12 16:43 # @Author : Lan # @File : 77.组合.py # @Software: PyCharm """ 给定两个整数 n 和 k，返回范围 [1, n] 中所有可能的 k 个数的组合。 你可以按 任何顺序 返回答案。 """ from typing import List class Solution: result = [] temp = [] def backtracking(self, n, k, start): if len(self.temp) == k: self.result.append(self.temp.copy()) else: for i in range(start, n - (k - len(self.temp)) + 2): self.temp.append(i) self.backtracking(n, k, i + 1) self.temp.pop() def combine(self, n: int, k: int) -> List[List[int]]: self.result = [] if n == k: return [[i for i in range(1, n + 1)]] self.backtracking(n, k, 1) return self.result print(Solution().combine(4, 2))
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某牛某客专栏文章爬虫 代码已脱敏，自行替换# @Time : 2023/10/8 14:43 # @Author : Lan # @File : niukespider.py # @Software: PyCharm import time import requests def get_category(catalog='10klpm'): url = f'https://www.lanol.cn.com/content/zhuanlan/index/catalog/{catalog}' return requests.get(url).json() c = """ <!doctype html> <html lang="en"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0"> <meta http-equiv="X-UA-Compatible" content="ie=edge"> <title>Document</title> </head> <body> {{content}} </body> </html> """ def get_content(catalog, entity): url = f'https://www.lanol.cn.com/content/zhuanlan/index/detail/{catalog}/{entity}?_={int(time.time() * 1000)}' return requests.get(url, headers={ 'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/117.0.0.0 Safari/537.36' }).json() if __name__ == '__main__': catalog = 'Gj5x2m' for i in get_category(catalog)['data']['catalog']: content = get_content(catalog, i['uuid'])['data'] with open('./docs/' + i['title'].replace('/', '-') + '.html', 'w', encoding='utf-8') as f: f.write(c.replace('{{content}}', content['content']))
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283. 移动零 移动零已解答简单相关标签相关企业提示给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。请注意 ，必须在不复制数组的情况下原地对数组进行操作。示例 1:输入: nums = [0,1,0,3,12]输出: [1,3,12,0,0]示例 2:输入: nums = [0]输出: [0]提示:1 <= nums.length <= 104-231 <= nums[i] <= 231 - 1进阶：你能尽量减少完成的操作次数吗？class Solution: def get_zero(self, slow, l_n, nums): for t in range(slow, l_n): if nums[t] == 0: return t def moveZeroes(self, nums: List[int]) -> None: fast, slow, l_n = 1, 0, len(nums), slow = self.get_zero(slow, l_n, nums) if slow is not None: while fast < l_n: if nums[fast] != 0 and fast > slow and slow is not None: nums[slow], nums[fast] = nums[fast], nums[slow] slow = self.get_zero(slow, l_n, nums) fast += 1
