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Commit 293d1a2

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ls1248659692zengli
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zengli
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Update leetcode_crawler.py
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‎.idea/leetcode.iml‎

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‎.idea/misc.xml‎

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‎README.md‎

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@@ -181,7 +181,8 @@ def sliding_window_template_with_examples(s, p):
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>* 剑指 Offer 04. 二维数组中的查找
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>* 剑指 Offer 21. 调整数组顺序使奇数位于偶数前面
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```python3
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# two pointers scenario, famous applications such as binary search, quick sort and sliding window.
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# two pointers scenario, famous applications such as binary search, quick sort and sliding window."""
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# --译:双指针场景,著名的应用包括二分查找、快速排序和滑动窗口(scenario:/səˈnærioʊ/ 场景;binary:/ˈbaɪnəri/ 二进制的;quicksort:/kwɪk sɔːrt/ 快速排序;sliding:/ˈslaɪdɪŋ/ 滑动的)
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'''
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Classification:

‎algorithm_templates/backtracking/backtracking.py‎

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# ("backtracks") as soon as it determines that the candidate cannot possibly be completed to a valid solution.
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#
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# backtracking vs. dfs
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# 译:回溯法与深度优先搜索
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# traverse in solution space vs. traverse in data structure space, pruned of dfs
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# 译:在解空间中遍历与在数据结构空间中遍历,深度优先搜索的剪枝
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def backtracking(self, data, candidate):

‎py_tricks/maxsplit.py‎

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#! /usr/bin/env python3
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"""split a string max times"""
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"""split a string max times -- 译:将字符串拆分最多次数(split:/spl?t/ 拆分)"""
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string = "a_b_c"
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print(string.split("_", 1))
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"""use maxsplit with arbitrarywhitespace"""
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"""use max split with arbitrary whitespace -- 译:使用按任意空白字符的最大拆分次数(arbitrary:/?ɑ?rb?treri/ 任意的;whitespace:/?wa?t.spe?s/ 空白字符 )"""
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s = "foo bar foobar foo"
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‎spider/leetcode_crawler.py‎

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@@ -142,7 +142,7 @@ def get_problems_describe(self, filters=None):
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print('sqlite ',IS_SUCCESS)
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while not IS_SUCCESS and req_retry<3:
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# try:
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# 休眠随机 1 - 2 秒,以免爬去频率过高被服务器禁掉
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# 爬虫休眠随机 1 - 2 秒,爬取频率过高会被服务器检测到访问频率过快而禁掉ip的问题
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time.sleep(random.randint(1, 5))
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req_retry+=1
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cursor = conn.cursor()
@@ -577,7 +577,6 @@ def generate_questions_submission(self, path, filters):
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print('Specified tag is: {}'.format(args.tags))
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leetcr.connect_mysql()
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# leetcr.get_problems_describe()
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# leetcr.get_ac_questions_submission_json(filters,skipsubm=True)
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if args_dict.get('code'):

‎spider/problems/242-valid-anagram/README.md‎

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return sorted(list(s))==sorted(list(t))
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```
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## 解法
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```python3
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class Solution:
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def isAnagram(self, s: str, t: str) -> bool:
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if not s or not t:
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return False
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if len(s) != len(t):
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return False
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#排序方式
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# return sorted(list(s)) == sorted(list(t))
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# 字典方式:
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# s 和 t 仅包含小写字母 ,这里也可以通过ord返回ASCII字符对应的整数直接定义数组的方式处理
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s_map = dict()
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for c in s:
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s_map[c] = s_map.get(c, 0) + 1
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for a in t:
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if a not in s_map:
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return False
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else:
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cnt = s_map.get(a, 0) - 1
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s_map[a] = cnt
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if cnt < 0:
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return False
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return True
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```
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[title]: https://leetcode-cn.com/problems/valid-anagram

‎spider/problems/454-4sum-ii/README.md‎

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**Difficulty:** Medium
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## 思路
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```python3
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from typing import List
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# leetcode submit region begin(Prohibit modification and deletion)
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class Solution:
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def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
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res = 0
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dic = {}
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# 计算前两个list对应的累计值字典:key:累加值,val:下标
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for i, a in enumerate(nums1):
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for j, b in enumerate(nums2):
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dic[a + b] = dic.get(a + b, 0) + 1
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# 遍历后两个数组计算,值是否存在
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for i, a in enumerate(nums3):
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for j, b in enumerate(nums4):
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c = 0 - a - b
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if c in dic and dic.get(c) > -1:
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# 累加时:注意,这里如果值相同,但是位置不同,算是多种组合,所以这里直接累加字典中该key的值即可
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res += dic.get(c)
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return res
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# leetcode submit region end(Prohibit modification and deletion)
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```
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[title]: https://leetcode-cn.com/problems/4sum-ii

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