原题是：
Given an array and a value, remove all instances of that value
in-placeand return the new length.
Do not allocate extra space for another array, you must do this by
modifying the input arrayin-placewith O(1) extra memory.
The order of elements can be changed. It doesn't matter what you leave beyond the new length.
Example:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
思路：
这个题与26题一样，都是比较经典的两指针思想。
i指针守住需要更新的位置，j指针去寻找符合条件的元素回来给i所在的位置。
注意这一题，由于自己设计的逻辑，需要返回的是i,而不再是i+1
代码是：
class Solution:
def removeElement(self, nums, val):
"""
:type nums: List[int]
:type val: int
:rtype: int
"""
i,j = 0,0
while j < len(nums):
if nums[j] != val:
nums[i] = nums[j]
i += 1
j += 1
else:
j += 1
return i
