LeetCode Notes_#27 Remove Element
Contents
题目
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.
思路和解答
思路
我想到了刚才看到的remove方法,这样做是不是不太好...
不对，remove只可以移除第一个匹配的项，所以不行
不用方法的话，那么就是使用直接循环，就是遇到指定的val就删除
解答
class Solution(object):
def removeElement(self, nums, val):
"""
:type nums: List[int]
:type val: int
:rtype: int
"""
while val in nums:
nums.remove(val)
return len(nums)
第一次写出faster than 100%...可以的
