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.
Example:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
注意点：
空间复杂度要求为O(1)
解题思路：
遍历列表，如果和列表中元素和目标元素相同，则从列表中删除此元素。删除后列表长度会改变，注意不要越界。
Pyhton实现：
class Solution:
def removeElement(self, nums, val):
"""
:type nums: List[int]
:type val: int
:rtype: int
"""
i = 0
while i < len(nums):
if nums[i] is val:
del nums[i]
else:
i +=1
return len(nums)
a = [3,2,2,3]
b = Solution()
print(b.removeElement(a,3)) # 2
