给定一个排序数组,你须要在原地删除重复出现的元素,使得每一个元素只出现一次,返回移除后数组的新长度。不要使用额外的数组空间,你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成。html
Remove Duplicates from Sorted Arraypython
# 解1,时间复杂度O(n²) class Solution: def removeDuplicates(self, nums: list) -> int: for i in range(len(nums)-1,0,-1): if nums[i] == nums[i-1]: nums.pop(i) return len(nums) # 解2,时间复杂度O(n) class Solution: def removeDuplicates(self, nums: list) -> int: i = 0 for j in range(0,len(nums)): if nums[i] != nums[j]: nums[i+1] = nums[j] i += 1 return i+1 if nums else 0
给定一个数组,将数组中的元素向右移动 k 个位置,其中 k 是非负数。
输入: [1,2,3,4,5,6,7] 和 k = 3
输出: [5,6,7,1,2,3,4]
解释:向右旋转 1 步: [7,1,2,3,4,5,6]
向右旋转 2 步: [6,7,1,2,3,4,5]
向右旋转 3 步: [5,6,7,1,2,3,4]git
Rotate Arraygithub
# 解1,数组拼接,空间复杂度O(n) class Solution: def rotate(self, nums: list, k: int) -> None: k %= len(nums) nums[:] = nums[-k:] + nums[:-k] return nums # 解2,3次翻转,空间复杂度O(1) class Solution: def rotate(self, nums: list, k: int) -> None: k, start, end = k % len(nums), 0, len(nums) - 1 self._reverse(nums, start, end - k) self._reverse(nums, end-k+1, end) self._reverse(nums, start, end) def _reverse(self, nums, start, end): while start < end: nums[start], nums[end] = nums[end], nums[start] start += 1 end -= 1
Web Automation: Selenium WebDriver and Python — Getting Started — Part 1web
简单的介绍了Web自动化,入门的文章chrome
打开cmder.exe时报错“failed to copy vendor/conemu.xml.default file to vendor/conemu-maximus5/conemu.xml access denied”
解决方法:原目录是“c:/program files/",建立一个新目录并将cmder文件夹复制进去
参考连接数组