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Xiaobian to share with you how LeetCode to remove elements, I hope you have something to gain after reading this article, let's discuss it together!

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Title Description

Given an array numbers and a value val, remove all elements equal to val in place and return the new length of the removed array. Only O (1) extra space is required, and the order of array elements can be changed. For example, enter [3,2,2,3], 3, and return 2.

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answer key

Idea: double pointer in turn to determine whether the elements in the list and val equal, equal to eliminate. With LeetCode brush question DAY 23: delete duplicate items in the sorted array similar situation, from left to right to look for attention to the impact of changes in the length of the array, the specific code is as follows, to avoid this impact makes the code simpler can be found from right to left, here temporarily do not show the code. class Solution: def removeElement(self, nums: List[int], val: int) -> int: if len(nums)==0: return i=0 while i < len(nums): if nums[i] == val: #Delete a value somewhere in the list nums.pop(i) else: i+=1 return len(nums)

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