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【LeetCode 27】Remove Element(Python)

2017-11-21 02:25 369 查看
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:

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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标签:  leetcode python