Leetcode日练笔记15 #27 # 26 Remove Element & Remove Duplicates from Sorted Array

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本文探讨了两种使用Python删除数组中特定元素和去重的方法。第一种方法利用`del`关键字，通过while循环找到并删除元素；第二种方法通过双指针策略，保持数组非降序的情况下删除重复项。这两种方法都要求原地修改数组并保持O(1)的额外空间复杂度。对于去除数组重复元素的解决方案，文章还对比了一种更高效的策略，即在遇到重复元素时不删除，而是跳过，最后返回指针位置作为结果。

#27 Remove Element

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The relative order of the elements may be changed.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

解题思路：

占了python del 功能的好处。一删完一个元素，下一个自动靠拢。所以就比较简单。

用while循环查找val在nums里的index，然后用del 删掉。最后输出nums的长度。

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        while val in nums:
            del_idx = nums.index(val)
            del nums[del_idx]
        return len(nums)

runtime:

#26 Remove Duplicates from Sorted Array

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

解题思路：

两个pointer，比较左右值，相等则删掉右值。不等则左标右标都右移一位。最后右标超出nums长度则停止循环。

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        left, right = 0, 1
        while right < len(nums):
            if nums[left] == nums[right]:
                del nums[right]
            else:
                left += 1
                right += 1
        return len(nums)

runtime:

看了一下65ms的solution：

这个思路相比之下，快在不删掉重复的值，而是直接不同的才挪坑改值。pointer的作用是保留unique的值。最后看pointer停留在idx的位置。确实步骤省了很多。瑞思拜。