【LeetCode 27】Remove Element(Python)
2017-11-21 02:25
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Given an array and a value, 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.
题目分析:给定一个数组和一个值。删掉所有出现在这个数组中的给定值,然后返回长度。
示例: nums=[1,3,2,3,4,2] val=3—————–结果:4
方法一:
思路:这道题和LeetCode26题基本类似。也是超级简单,没什么好说的,有问题可以参照我上一篇博客(【LeetCode 26】Remove Duplicates from Sorted Array)
优点:时间复杂度O(1)
代码:
方法二:
缺点:时间复杂度O(n)
代码:
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.
题目分析:给定一个数组和一个值。删掉所有出现在这个数组中的给定值,然后返回长度。
示例: nums=[1,3,2,3,4,2] val=3—————–结果:4
方法一:
思路:这道题和LeetCode26题基本类似。也是超级简单,没什么好说的,有问题可以参照我上一篇博客(【LeetCode 26】Remove Duplicates from Sorted Array)
优点:时间复杂度O(1)
代码:
class Solution: def removeElement(self, nums, val): """ :type nums: List[int] :type val: int :rtype: int """ while val in nums: nums.remove(val) return len(nums)
方法二:
缺点:时间复杂度O(n)
代码:
class Solution: def removeElement(self, nums, val): """ :type nums: List[int] :type val: int :rtype: int """ a=0 for i in nums: if i!=val: nums[a]=i a+=1 return a
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