leetcode 027 移除元素(Remove Element) python3 最简代码

本文提供了一种高效的方法来解决LeetCode上的移除元素问题。通过在原地修改输入数组并返回新长度,避免了额外空间的使用,遵循O(1)额外内存的要求。文章详细介绍了两种方法,包括一种利用while循环去除指定值的简洁解决方案。

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所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正,讨论交流。
'''
Given an array nums and a value val, remove all instances of that value in-place and return the new length.

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

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example 1:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
It doesn't matter what you leave beyond the returned length.

Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyond the returned length.
Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

'''



class Solution:
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        # Approach 1: 同时得到val的个数,以及按照原顺序的新列表,参考 #283 题 https://blog.youkuaiyun.com/huhehaotechangsha/article/details/80533705
        # number = 0
        # for i,num in enumerate(nums):
        #     if num != val:
        #         nums[number], nums[i] = nums[i] ,nums[number]
        #         number += 1
        # return number

        #Approach 2:  不用for循环
        # Time: O(n)
        # Space:O(1)
        while val in nums:
            nums.remove(val)
        return len(nums)
所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正,讨论交流。
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